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𝑐á𝑐 𝑏à𝑖 ℎệ phương trình ℎ𝑎𝑦 𝑣à 𝑘ℎó
(Nguyễn Trường Phát TP.HCM)
𝑥 + 𝑥 √𝑥 − 3√𝑥 − 2𝑥𝑦 + 6𝑦 − 3 = 0(1)
𝑏à𝑖 1: {
3
√𝑥 + 2𝑦 + 1 + √𝑥 − 2𝑦 + 2 = 3(2)
𝑔𝑖ả𝑖
𝑥 + 2𝑦 + 1 ≥ 0
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ {
𝑥≥0
3

2

𝑝𝑡(1): (√𝑥) + (√𝑥) − 3√𝑥 − 3 − 2𝑥𝑦 + 6𝑦 = 0
𝑥(√𝑥 + 1) − 3(√𝑥 + 1) − 2𝑦(𝑥 − 3) = 0
(𝑥 − 3)(√𝑥 + 1) − 2𝑦(𝑥 − 3) = 0
𝑣ậ𝑦 𝑥 = 3 ℎ𝑎𝑦 √𝑥 + 1 − 2𝑦 = 0
(∗∗)𝑣ớ𝑖 𝑥 = 3 𝑡ℎế 𝑣à𝑜 (2)𝑡ℎì 𝑡𝑎 𝑡ì𝑚 đượ𝑐 𝑦 = 16 𝑣à 𝑦 = −

3

2

𝑡ℎế 𝑥 = 3 𝑣à 𝑦 = 16 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎
𝑡ℎế 𝑥 = 3 𝑣à 𝑦 = −

3
𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑐ũ𝑛𝑔 𝑡ℎõ𝑎
2

(∗)𝑣ớ𝑖 √𝑥 − 2𝑦 + 1 = 0 𝑡ℎế 𝑣à𝑜 (2)
3

√𝑥 + √𝑥 + 2 + √𝑥 − √𝑥 + 1 = 3(𝑙𝑖ê𝑛 ℎợ𝑝 𝑣ớ𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = 1)
𝑡ℎế 𝑣à𝑜 (∗)𝑡𝑎 𝑡ì𝑚 đượ𝑐 𝑦 = 1
𝑣ớ𝑖 𝑥 = 1, 𝑦 = 1 𝑡ℎế 𝑣à𝑜 (1) 𝑐ũ𝑛𝑔 𝑡ℎõ𝑎
3
𝑣ậ𝑦 ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑛ℎư 𝑠𝑎𝑢: (1,1); (3,16); (3, − )
2


2𝑥 + 1 = 𝑦 2 + 4𝑦√𝑥(1)
𝑏à𝑖 2: {
𝑦 + √𝑥 + 1 = 𝑦√𝑥(2)
𝑔𝑖ả𝑖
đ𝑘 để ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à 𝑥 ≥ 0
để ý 𝑝𝑡(2): 𝑦 + 1 = 𝑦√𝑥 − √𝑥
𝑙ấ𝑦(2)𝑡ℎế 𝑣à𝑜 (1) 𝑡𝑎 𝑐ó đượ𝑐: 2𝑥 − 4𝑦√𝑥 = (𝑦 + 1)(𝑦 − 1)
= (𝑦√𝑥 − √𝑥)(𝑦 − 1)
√𝑥 = 0 ℎ𝑎𝑦 2√𝑥 − 4𝑦 = 𝑦 2 − 2𝑦 + 1
√𝑥 = 0 ℎ𝑎𝑦 2√𝑥 = 𝑦 2 + 2𝑦 + 1
𝑣ớ𝑖 𝑥 = 0 𝑡ℎì 𝑦 = ±1 𝑡ℎế 𝑣à𝑜 (2)𝑡𝑎 𝑛ℎậ𝑛 𝑥 = 0, 𝑦 = −1
𝑦 2 + 2𝑦 + 1
𝑣ớ𝑖 √𝑥 =
𝑡ℎế 𝑣à𝑜 (2)
2
𝑦 2 + 2𝑦 + 1
𝑦 2 + 2𝑦 + 1
𝑦+
+1 = 𝑦(
)


2
2
𝑦 2 + 4𝑦 + 3 = 𝑦 3 + 2𝑦 2 + 𝑦
𝑦 = −1 𝑡ℎì 𝑥 = 0 ℎ𝑎𝑦 𝑦 = ±√3, 𝑥 = 7 ± 4√3
𝑣ớ𝑖 𝑦 = −1 𝑣à 𝑥 = 0 𝑡ℎế 𝑛𝑔ượ𝑐 𝑙ạ𝑖 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎
𝑣ớ𝑖 𝑦 = √3 𝑣à 𝑥 = 7 + 4√3 𝑡ℎế 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎
𝑣ớ𝑖 𝑦 = −√3 𝑣à 𝑥 = 7 − 4√3 𝑡ℎế 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑘ℎô𝑛𝑔 𝑡ℎõ𝑎
𝑣ậ𝑦 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑙ầ𝑛 𝑙ượ𝑡 𝑙à: (0, −1); (√3, 7 + 4√3)
√3𝑥 − 1 + 4(2𝑥 + 1) = √𝑦 − 1 + 3𝑦(1)
𝑏à𝑖 3: {
(𝑥 + 𝑦)(2𝑥 − 𝑦) + 6𝑥 + 3𝑦 + 4 = 0(2)


𝑔𝑖ả𝑖
1
đ𝑘 để ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ {
3
𝑦≥1
𝑥≥

𝑝𝑡(2): 2𝑥 2 − 𝑥𝑦 + 2𝑥𝑦 − 𝑦 2 + 6𝑥 + 3𝑦 + 4 = 0
2𝑥 2 + 𝑥𝑦 − 𝑦 2 + 6𝑥 + 3𝑦 + 4 = 0
2𝑥 2 + 𝑥(𝑦 + 6) − 𝑦 2 + 3𝑦 + 4 = 0
𝑑𝑒𝑛𝑡𝑎 = 𝑦 2 + 12𝑦 + 36 − 8(−𝑦 2 + 3𝑦 + 4) = 9𝑦 2 − 12𝑦 + 4 = (3𝑦 − 2)2
≥0
−𝑦 − 6 + 3𝑦 − 2 2𝑦 − 8 𝑦 − 4
𝑥=
=
=
4
4
2
𝑣ậ𝑦 [
−𝑦 − 6 − 3𝑦 + 2 −4𝑦 − 4
𝑥=
=
= −𝑦 − 1
4
4
𝑣ớ𝑖 𝑥 =

𝑦−4
𝑡ℎế 𝑣à𝑜 (1)𝑡𝑎 đượ𝑐 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à 𝑦 = 12, 𝑥 = 4
2

𝑣ớ𝑖 𝑥 + 𝑦 = −1 𝑡ℎì 𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚 𝑑𝑜 𝑥 + 𝑦 > 0
𝑣ậ𝑦 ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ (4,12)
𝑏à𝑖 4: {

(𝑥 2 + 2)√𝑦 + 1 − (𝑦 + 2)√𝑥 2 + 1 = 0(1)
𝑥 4 − 𝑦 2 − 3𝑥 2 𝑦 − 5𝑦 + 8 = 0(2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: 𝑦 ≥ −1

𝑝𝑡(1): (𝑥 4 + 4𝑥 2 + 4)(𝑦 + 1) = (𝑦 2 + 4𝑦 + 4)(𝑥 2 + 1)
𝑥 4 𝑦 + 𝑥 4 + 4𝑥 2 𝑦 + 4𝑥 2 + 4𝑦 + 4 = 𝑦 2 𝑥 2 + 𝑦 2 + 4𝑦𝑥 2 + 4𝑦 + 4𝑥 2 + 4
𝑥 4𝑦 + 𝑥 4 = 𝑥 2𝑦2 + 𝑦2
𝑥 2 𝑦(𝑥 2 − 𝑦) = (𝑦 − 𝑥 2 )(𝑦 + 𝑥 2 )


𝑣ậ𝑦 𝑥 2 = 𝑦 ℎ𝑎𝑦 𝑥 2 𝑦 = −(𝑦 + 𝑥 2 )
𝑣ớ𝑖 𝑥 2 = 𝑦 𝑡ℎế 𝑣à𝑜 (2): −3𝑦 2 − 5𝑦 + 8 = 0[

𝑦=1
8
𝑦=−
3

𝑥2
𝑣ớ𝑖 𝑥 𝑦 + 𝑦 = −𝑥 => 𝑦 = − 2
𝑡ℎế 𝑣à𝑜 (2)
𝑥 +1
2

2

2

𝑥2
𝑥2
𝑥2
2
𝑥 −( 2
) + 3𝑥 ( 2
) + 5( 2
)+8= 0
𝑥 +1
𝑥 +1
𝑥 +1
4

𝑥 4 (𝑥 2 + 1)2 − 𝑥 4 + 3𝑥 4 (𝑥 2 + 1) + 5𝑥 2 (𝑥 2 + 1) + 8(𝑥 2 + 1)2 = 0
𝑥 4 (𝑥 4 + 2𝑥 2 + 1) − 𝑥 4 + 3𝑥 6 + 3𝑥 4 + 5𝑥 4 + 5𝑥 2 + 8(𝑥 2 + 1)2 = 0
𝑥 8 + 2𝑥 6 + 3𝑥 6 + 8𝑥 4 + 5𝑥 2 + 8(𝑥 4 + 2𝑥 2 + 1) = 0
𝑥 8 + 5𝑥 6 + 16𝑥 4 + 21𝑥 2 + 8
= 0 (𝑝𝑡 𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚 𝑑𝑜 𝑥 8 + 5𝑥 6 + 16𝑥 4 + 21𝑥 2 + 8 > 0)

2𝑥 2 + 𝑥 + √𝑥 + 2 = 2𝑦 2 + 𝑦 + √2𝑦 + 1(1)
𝑏à𝑖 5: {
𝑥 2 + 2𝑦 2 − 2𝑥 + 𝑦 − 2 = 0(2)
𝑔𝑖ả𝑖
𝑥 ≥ −2
1
đ𝑘 để 𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ {
𝑦≥−
2
𝑝𝑡(1) − 𝑝𝑡(2): 𝑥 2 − 2𝑦 2 + 3𝑥 − 𝑦 + 2 + √𝑥 + 2 = 2𝑦 2 + 𝑦 + √2𝑦 + 1
(𝑥 + 1)2 + (𝑥 + 1) + √(𝑥 + 1) + 1 = (2𝑦)2 + 2𝑦 + √2𝑦 + 1
𝑥é𝑡 𝑓(𝑡) = 𝑡 2 + 𝑡 + √𝑡 + 1 𝑣ậ𝑦 𝑓 ′ (𝑡) = 2𝑡 + 1 +

1
2 √𝑡 + 1

𝑡ứ𝑐 𝑓(𝑥 + 1) = 𝑓(2𝑦)𝑣ậ𝑦 𝑥 + 1 = 2𝑦(3)

>0


𝑙ấ𝑦 (3)𝑡ℎế (2): (2𝑦 − 1)2 + 2𝑦 2 − 2(2𝑦 − 1) + 𝑦 − 2 = 0
4𝑦 2 − 4𝑦 + 1 + 2𝑦 2 − 4𝑦 + 2 + 𝑦 − 2 = 0
6𝑦 2 − 7𝑦 + 1 = 0 [

𝑦=1
1
𝑦=
6

𝑣ớ𝑖 𝑦 = 1 𝑣ậ𝑦 𝑥 = 1
𝑣ớ𝑖 𝑦 =

1
2
𝑣ậ𝑦 𝑥 = −
6
3

𝑥 3 𝑦√𝑥 + 𝑥 3 𝑦 2 = 2𝑥 4 √𝑥 + 2𝑥 4 𝑦(1)

𝑏à𝑖 6: {
𝑦√𝑥 (√2𝑥 2 − 6 − 1) = √5(2𝑥 2 − 6)(2)
𝑔𝑖ả𝑖
𝑝𝑡(1): 𝑥 3 𝑦(√𝑥 + 𝑦) = 2𝑥 4 (√𝑥 + 𝑦)
𝑣ậ𝑦 √𝑥 = −𝑦 ℎ𝑎𝑦 𝑥 = 0 ℎ𝑎𝑦 𝑦 = 2𝑥

𝑥 3 − 3𝑦 3 − 3𝑥 2 𝑦 + 𝑥𝑦 2 + 𝑥 = 3𝑦(1)
𝑏à𝑖 7:

3

3

2

3𝑥 + 36𝑦 − 1 = 𝑥 √27𝑦 3 +
{

2𝑥 + 1
(2)
𝑥

𝑔𝑖ả𝑖
𝑝𝑡(1): (𝑥 − 3𝑦)(𝑥 2 + 𝑦 2 ) = −(𝑥 − 3𝑦)
𝑣ậ𝑦 𝑥 = 3𝑦 ℎ𝑎𝑦 𝑥 2 + 𝑦 2 = −1(𝑝𝑡 𝑣𝑛 𝑑𝑜 𝑥 2 + 𝑦 2 ≥ 0)
𝑣ớ𝑖 𝑥 = 3𝑦 => 𝑦 =

𝑥
𝑡ℎế (2):
3

1
3𝑥 + 4𝑥 − 1 = 𝑥 √𝑥 3 + 2 + (𝑙𝑖ê𝑛 ℎợ𝑝 )
𝑥
3

2

3


2𝑥(𝑥 2 + 3) − 𝑦(𝑦 2 + 3) = 3𝑥𝑦(𝑥 − 𝑦)(1)
𝑏à𝑖 8: {
(𝑥 2 − 2)2 = 4(2 − 𝑦)(2)
𝑔𝑖ả𝑖
𝑝𝑡(1)2𝑥 3 + 6𝑥 − 𝑦 3 − 3𝑦 = 3𝑥 2 𝑦 − 3𝑥𝑦 2
2𝑥 3 − 3𝑥 2 𝑦 + 3𝑥𝑦 2 − 𝑦 3 + 6𝑥 − 3𝑦 = 0
(2𝑥 − 𝑦)(𝑥 2 − 𝑥𝑦 + 𝑦 2 ) + 3(2𝑥 − 𝑦) = 0
2𝑥 = 𝑦 ℎ𝑎𝑦 𝑥 2 − 𝑥𝑦 + 𝑦 2 = −3 (𝑝𝑡 𝑣𝑛)
𝑣ờ𝑖 2𝑥 = 𝑦 𝑡ℎế 𝑣à𝑜 (2):
(𝑥 2 − 2)2 = 4(2 − 2𝑥)
𝑥 4 − 4𝑥 2 + 4 = 8 − 8𝑥
(𝑥 2 + 2𝑥 − 2)(… ) = 0 => 𝑥 = −1 ± √3
𝑏à𝑖 9: {

√𝑥 2 + 7 + √−2 − 𝑥 + √−1 − 10𝑦 = 8(1)
𝑥√𝑦 2 − 4𝑦 + 5 + √𝑥 2 + 1 = 𝑦√𝑥 2 + 1 − √𝑦 2 − 4𝑦 + 5(2)
(đã 𝑔𝑖ả𝑖 𝑟ồ𝑖 ‼!)
2(2𝑥 − 1)3 + 2𝑥 = (2𝑦 − 1)√𝑦 − 1 + 1(1)
𝑏à𝑖 10: {
√4𝑥 − 2 + √2𝑦 − 2 = 2√2(2)
𝑔𝑖ả𝑖
𝑦≥1
1
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ {
𝑥≥
2
𝑝𝑡(1): 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ℎ𝑜 2
1
1
8𝑥 3 − 12𝑥 2 + 6𝑥 − 1 + 𝑥 = (𝑦 − ) √𝑦 − 1 +
2
2


8𝑥 3 − 12𝑥 2 + 6𝑥 −
(2𝑥 − 1)3 +

3
3
√𝑦 − 1
= (√𝑦 − 1) +
2
2

2𝑥 − 1
3
√𝑦 − 1
= (√𝑦 − 1) +
2
2

𝑥é𝑡 𝑓(𝑡) = 𝑡 3 +

𝑡
1
=> 𝑓 ′ (𝑡) = 3𝑡 2 + > 0
2
2

𝑣ậ𝑦 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓(2𝑥 − 1) = 𝑓(√𝑦 − 1)
𝑣ậ𝑦 2𝑥 − 1 = √𝑦 − 1(𝑡ℎế 𝑣à𝑜 (2))
4𝑥 2 − 4𝑥 + 1 = 𝑦 − 1
𝑦 = 4𝑥 2 − 4𝑥 + 2
√4𝑥 − 2 + √2(4𝑥 2 − 4𝑥 + 2) − 2 = 2√2
√4𝑥 − 2 + √8𝑥 2 − 8𝑥 + 2 = 2√2(𝑙𝑖ê𝑛 ℎợ𝑝 𝑥
= 1 𝑙ư𝑢 ý 𝑠𝑎𝑢 𝑘ℎ𝑖 𝑙𝑖ê𝑛 ℎợ𝑝 𝑛ℎớ 𝑏ấ𝑚 𝑛𝑔ℎ𝑖ệ𝑚 𝑙ạ𝑖 𝑙ầ𝑛 𝑛ữ𝑎 𝑡𝑟á𝑛ℎ 𝑛𝑔ℎ𝑖ệ𝑚 𝑘é𝑝)
𝑥 = 1 𝑡ℎì 𝑦 = 2
𝑣ậ𝑦 ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ (1,2)
(𝑥 − 𝑦)(𝑥 + 𝑦 + 𝑦 2 ) = 𝑥(𝑦 + 1)(1)
𝑏à𝑖 11: {
(𝑦 + 2)2
√𝑥 3 + 4𝑥 = 1 +
(2)
3
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ 𝑥 ≥ 0
𝑝𝑡(1): 𝑥 2 + 𝑥𝑦 + 𝑥𝑦 2 − 𝑥𝑦 − 𝑦 2 − 𝑦 3 = 𝑥𝑦 + 𝑥
𝑥 2 + 𝑥𝑦 2 − 𝑦 2 − 𝑦 3 = 𝑥𝑦 + 𝑥
𝑥(𝑥 + 𝑦 2 ) = 𝑥(𝑦 + 1) + 𝑦 2 (𝑦 + 1)
𝑥(𝑥 + 𝑦 2 ) = (𝑥 + 𝑦 2 )(𝑦 + 1)


𝑣ậ𝑦 𝑥 + 𝑦 2 = 0 ℎ𝑎𝑦 𝑥 = 𝑦 + 1
𝑣ớ𝑖 𝑥 + 𝑦 2 = 0 (𝑝𝑡𝑣𝑛)𝑑𝑜 𝑥 + 𝑦 2 > 0
𝑣ớ𝑖 𝑦 = 𝑥 − 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)
√𝑥 3

(𝑥 + 1)2
𝑥 2 + 2𝑥 + 1
+ 4𝑥 = 1 +
=1+
3
3

3√𝑥 3 + 4𝑥 = 𝑥 2 + 2𝑥 + 4(𝑙𝑖ê𝑛 ℎợ𝑝 𝑣ớ𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = 2)
𝑣ớ𝑖 𝑥 = 2 𝑡ℎì 𝑦 = 1
𝑣ậ𝑦 ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: (2,1)
𝑥 3 + 𝑦 3 − 6𝑥 2 + 15𝑥 + 3𝑦 − 14 = 0(1)
𝑏à𝑖 12: {
4
4
√𝑥 + √𝑦 + √𝑥 + √𝑦 = 4(2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à 𝑥, 𝑦 ≥ 0
á𝑝 𝑑ụ𝑛𝑔 𝐵Đ𝑇 𝑏𝑢ℎ𝑖𝑎, 𝑑ấ𝑢=𝑥ả𝑦 𝑟𝑎 𝑘ℎ𝑖 𝑥 = 𝑦 = 1
𝑡ℎế 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎 𝑣ậ𝑦
ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ (1,1)
1 + (𝑥𝑦)3 = 19𝑥 3 (1)
𝑏à𝑖 13: {
𝑦 + 𝑥𝑦 2 = −6𝑥 2 (2)
𝑔𝑖ả𝑖
𝑇𝐻1: 𝑥 = 0 𝑡ℎế 𝑣à𝑜 (1) 𝑝𝑡𝑣𝑛
𝑇𝐻2: 𝑥 ≠ 0 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ủ𝑎 𝑝𝑡(1)𝑐ℎ𝑜 𝑥 3 :
1
+ 𝑦 3 = 19
3
𝑥
𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ủ𝑎 𝑝𝑡(2)𝑐ℎ𝑜 𝑥 2 :


𝑦 𝑦2
+
= −6
𝑥2 𝑥
𝑡𝑎 𝑐ó − 6. 𝑝𝑡(1) − 19. 𝑝𝑡(2) = 0
6
19𝑦 19𝑦 2
3
− 3 − 6𝑦 − 2 −
=0
𝑥
𝑥
𝑥
3
2𝑥
1
[
𝑦=−
𝑥
2
𝑦=−
3𝑥
𝑦=−

16(𝑥𝑦)3 − 9𝑦 3 = (2𝑥𝑦 − 𝑦)(4𝑥𝑦 2 + 3)(1)
𝑏à𝑖 14: {
4𝑥 2 𝑦 2 + 2𝑥𝑦 2 + 𝑦 2 = 3(2)
𝑔𝑖ả𝑖
𝑡ℎế (2) 𝑣à𝑜 (1): 16𝑥 3 𝑦 3 − 9𝑦 3 = 𝑦 3 (2𝑥 − 1)(4𝑥 2 + 6𝑥 + 1)
𝑣ậ𝑦 𝑦 = 0 ℎ𝑎𝑦 16𝑥 3 − 9 = 8𝑥 3 + 12𝑥 2 + 2𝑥 − 4𝑥 2 − 6𝑥 − 1
𝑦 = 0 ℎ𝑎𝑦 8𝑥 3 − 8𝑥 2 + 4𝑥 − 8 = 0
2 + 6𝑦 =
𝑏à𝑖 15:

𝑥
− √𝑥 − 2𝑦(1)
𝑦

√𝑥 + √𝑥 − 2𝑦 = 𝑥 + 3𝑦 − 2(2)
{
𝑔𝑖ả𝑖
𝑦≠0
𝑥 − 2𝑦 ≥ 0

đ𝑘 để ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ {
𝑥 + √𝑥 − 2𝑦 ≥ 0
𝑝𝑡(1): 2𝑦 + 6𝑦 2 = 𝑥 − 𝑦√𝑥 − 2𝑦
đặ𝑡 𝑡ℎử ℎệ 𝑠ố ∶ 𝑥 − 2𝑦 + 𝑎𝑏 + (𝑎 + 𝑏)√𝑥 − 2𝑦 = 0


{

𝑥 − 2𝑦 + 𝑎𝑏 = 𝑥 − 2𝑦 − 6𝑦 2
𝑎 + 𝑏 = −𝑦
(−𝑦)𝑏 − 𝑏 2 + 6𝑦 2 = 0(∗)
{
𝑎 = (−𝑦) − 𝑏

𝑑𝑒𝑛𝑡𝑎(∗) = 𝑦 2 + 24𝑦 2 = (5𝑦)2 ≥ 0
𝑦 − 5𝑦
√𝑥 − 2𝑦 = −2 = 2𝑦
[
𝑦 + 5𝑦
√𝑥 − 2𝑦 = −2 = −3𝑦

(𝑥 + 𝑦)(𝑥 + 4𝑦 2 + 𝑦) + 3𝑦 4 = 0(1)
𝑏à𝑖 16: {
√𝑥 + 2𝑦 2 + 1 − 𝑦 2 + 𝑦 + 1 = 0(2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ 𝑥 + 2𝑦 2 ≥ −1
để ý 𝑝𝑡(1): 𝑥 2 + 4𝑥𝑦 2 + 2𝑥𝑦 + 4𝑦 3 + 𝑦 2 + 3𝑦 4 = 0
𝑥 2 + 𝑥(4𝑦 2 + 2𝑦) + 4𝑦 3 + 𝑦 2 + 3𝑦 4 = 0
𝑑𝑒𝑛𝑡𝑎 = 16𝑦 4 + 16𝑦 3 + 4𝑦 2 − 16𝑦 3 − 4𝑦 2 − 12𝑦 4 = 4𝑦 4 ≥ 0
−4𝑦 2 − 2𝑦 − 2𝑦 2
= −3𝑦 2 − 𝑦
2
[
−4𝑦 2 − 2𝑦 + 2𝑦 2
𝑥=
= −𝑦 2 − 𝑦
2
𝑥=

(∗)𝑣ớ𝑖 𝑥 = −3𝑦 2 − 𝑦 𝑡ℎế 𝑣à𝑜 (2):
√−𝑦 2 − 𝑦 + 1 − 𝑦 2 + 𝑦 + 1 = 0( 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑦 = −1)
𝑣ậ𝑦 𝑦 = −1 𝑡ℎì 𝑥 = −2 𝑡ℎế 𝑣à𝑜 đ𝑘 𝑡ℎấ𝑦 𝑡ℎõ𝑎
(∗∗)𝑣ớ𝑖 𝑥 = −𝑦 2 − 𝑦 𝑡ℎế 𝑣à𝑜 (2):
√𝑦 2 − 𝑦 + 1 − 𝑦 2 + 𝑦 + 1 = 0


𝑦2 − 𝑦 + 1 ≥ 0
𝑏ì𝑛ℎ 𝑝ℎươ𝑛𝑔 2 𝑣ế 𝑣ớ𝑖 đ𝑘 𝑙à ∶ { 2
𝑦 −𝑦−1≥0
𝑦 2 − 𝑦 + 1 = 𝑦 4 + 𝑦 2 + 1 − 2𝑦 3 + 2𝑦 − 2𝑦 2
𝑦=0
𝑦 − 2𝑦 − 2𝑦 + 3𝑦 = 0 [ 𝑦 = 1
1 ± √13
𝑦=
2
4

3

2

𝑥 3 − 3𝑥𝑦 + 3𝑥 = 6(1)

𝑏à𝑖 17: { 2
𝑥 √𝑦 − 1 − 2𝑥𝑦 = 1 − 2𝑥(2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: 𝑦 ≥ 1
𝑝𝑡(2): 6𝑥 2 √𝑦 − 1 − 12𝑥𝑦 + 12𝑥 = 6
𝑥 3 − 6𝑥 2 √𝑦 − 1 + 9𝑥𝑦 − 9𝑥 = 0
𝑥 = 0 ℎ𝑎𝑦 𝑥 2 − 6𝑥√𝑦 − 1 + 9(𝑦 − 1) = 0(𝑝𝑡𝑣𝑛)
(∗)𝑣ớ𝑖 𝑥 = 0 𝑡ℎế 𝑣à𝑜 (2)𝑡𝑎 𝑐ó 𝑝𝑡𝑣𝑛
2

(𝑥 − 3√𝑦 − 1) = 0
(∗)𝑣ớ𝑖 𝑥 = 3√𝑦 − 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2):
𝑥 2 = 9𝑦 − 9 (đ𝑘 𝑥 ≥ 0)
(𝑥 2 + 9)𝑥
𝑥 −
+ 3𝑥 = 6
3
3

3

𝑥 3 = 9 => 𝑥 = √9
𝑥 3 + 3𝑥 2 + 4𝑥 + 2 = (𝑦 + 2)√𝑦 + 1(1)
𝑏à𝑖 18: { 2
𝑥 + 𝑥𝑦 − 16𝑥 − 2𝑦 2 + 22𝑦 − 36 = 0(2)


𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ 𝑦 ≥ −1
3

𝑝𝑡(1): (𝑥 3 + 3𝑥 2 + 3𝑥 + 1) + (𝑥 + 1) = (√𝑦 + 1) + √𝑦 + 1
3

(𝑥 + 1)3 + (𝑥 + 1) = (√𝑦 + 1) + √𝑦 + 1
𝑥é𝑡 𝑓(𝑡) = 𝑡 3 + 𝑡 𝑣ậ𝑦 𝑡ℎì 𝑓 ′ (𝑡) = 3𝑡 2 + 1 > 0
𝑡ứ𝑐 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓(𝑥 + 1) = 𝑓(√𝑦 + 1)
𝑣ậ𝑦 𝑥 + 1 = √𝑦 + 1(đế𝑛 đâ𝑦 𝑡ℎì 𝑐á𝑐 𝑏ạ𝑛 𝑐ó 𝑡ℎể 𝑡ự 𝑙à𝑚 đượ𝑐 𝑟ò𝑖 )
𝑥 7 + 𝑥𝑦 6 = 𝑦14 + 𝑦 8 (1)

𝑏à𝑖 19: { 3
√𝑦 + 6 + 𝑥 = 7 − √𝑦 − 1(2)
𝑔𝑖ả𝑖
𝑝𝑡(1): 𝑇𝐻1 𝑔𝑖ả 𝑠ử 𝑦 = 0 𝑡ℎì 𝑥 = 0 𝑡ℎế 𝑣à𝑜 (2)𝑡𝑎 𝑡ℎấ𝑦 𝑘ℎô𝑛𝑔 𝑡ℎõ𝑎
𝑇𝐻2: 𝑔𝑖ả 𝑠ử 𝑦 ≠ 0 𝑡ℎì 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ủ𝑎 𝑝𝑡(1) 𝑐ℎ𝑜 𝑦 7
𝑥 7 𝑥
( ) + = 𝑦7 + 𝑦
𝑦
𝑦
𝑘ℎ𝑖 đó 𝑡𝑎 𝑥é𝑡 𝑓(𝑡) = 𝑡 7 + 𝑡 𝑣ậ𝑦 𝑓 ′ (𝑡) = 7𝑡 6 + 1 > 0
𝑥
𝑡ứ𝑐 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓 ( ) = 𝑓(𝑦)𝑣ậ𝑦 𝑥
𝑦
2 (𝑡ℎế
=𝑦
𝑣à𝑜 𝑝𝑡(2)𝑡í𝑛ℎ 𝑏ì𝑛ℎ 𝑡ℎườ𝑛𝑔)
𝑥
121
− 272 (1)
𝑏à𝑖 20: {
9
2
2
𝑥 + 𝑦 + 𝑥𝑦 − 3𝑥 − 4𝑦 + 4 = 0(2)

𝑥 2 + 2𝑥 =

𝑔𝑖ả𝑖
𝑝𝑡(1): 𝑦 + 𝑦(𝑥 − 4) + 𝑥 2 − 3𝑥 + 4 = 0
2


𝑑𝑒𝑛𝑡𝑎 = 𝑥 2 − 8𝑥 + 16 − 4𝑥 2 + 12𝑥 − 16 = 0 ≤ 𝑥 ≤
4
4 2
4
3.2
𝑝𝑡(2): 𝑥 + 2𝑥 +
≤ ( ) + 2 ( ) + 27
3
3
4
4
𝑣ậ𝑦 𝑑ấ𝑢 𝑏ằ𝑛𝑔 𝑥ả𝑦 𝑟𝑎 𝑥 = 𝑛ê𝑛 𝑦 =
3
3

4
3

𝑥
272

2

𝑥𝑦 2 (√𝑥 2 + 1 + 1) = 3√𝑦 2 + 9 + 3𝑦(1)

𝑏à𝑖 23: {
(3𝑥 − 1)√𝑥 2 𝑦 + 𝑥𝑦 − 5 − 4𝑥 3 + 3𝑥 3 𝑦 − 7𝑥 = 0(2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 𝑥𝑦(𝑥 + 1) ≥ 5
𝑇𝐻1: 𝑥é𝑡 𝑦 2 = 0
𝑇𝐻2: 𝑦 2 ≠ 0 𝑛ê𝑛 𝑡𝑎 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ủ𝑎 𝑝𝑡(1) 𝑙à 𝑦 2 :
𝑥 √𝑥 2 + 1 + 𝑥 =

3
3
2+9+
√𝑦
𝑦2
𝑦

2
4
3
3
3
√𝑥 4 + 𝑥 2 + 𝑥 = √( ) + ( ) +
𝑦
𝑦
𝑦

𝑥é𝑡 𝑓(𝑡) =

√𝑡 4

+

𝑡2

+ 𝑡 𝑣ậ𝑦 𝑓

′ (𝑡)

=

4𝑡 3 + 2𝑡
2√𝑡 4 + 𝑡 2

+1>0

3
𝑣ậ𝑦 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓(𝑥) = 𝑓 ( ) (𝑡ừ đâ𝑦 𝑡ℎế 𝑣à𝑜 (2))
𝑦

𝑥𝑦(𝑥 + 𝑦) = 2(1)
𝑏à𝑖 22: { 3
𝑦 + 𝑦 + 6 = 7𝑥 3 + 𝑥(2)
𝑔𝑖ả𝑖
3𝑥 2 𝑦 + 3𝑥𝑦 2 = 6
{ 3
7𝑥 + 𝑥 − 𝑦 3 − 𝑦 = 6


−7𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 − 𝑥 + 𝑦 = 0
−(𝑥 − 𝑦)(7𝑥 2 + 4𝑥𝑦 + 𝑦 2 ) − (𝑥 − 𝑦) = 0
𝑣ậ𝑦 𝑥 = 𝑦 ℎ𝑎𝑦 7𝑥 2 + 4𝑥𝑦 + 𝑦 2 + 1 = 0 (𝑝𝑡𝑣𝑛 𝑑𝑜 7𝑥 2 + 4𝑥𝑦 + 𝑦 2 + 1 > 0)
𝑥2 + 𝑦2
𝑥 2 + 𝑥𝑦 + 𝑦 2

+√
= 𝑥 + 𝑦(1)
𝑏à𝑖 23:
2
3
{ 𝑥√2𝑥𝑦 + 5𝑥 + 3 = 4𝑥𝑦 − 5𝑥 − 3(2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 2𝑥𝑦 + 5𝑥 + 3 ≥ 0
(𝑥 + 𝑦)2 − 2𝑥𝑦
(𝑥 + 𝑦)2 − 𝑥𝑦


𝑝𝑡(1):
+
=𝑥+𝑦
2
3
𝑇𝐻1: 𝑥 + 𝑦 = 0 𝑡ứ𝑐 𝑥 = −𝑦 𝑡ℎế 𝑣à𝑜 (2):
𝑥√−2𝑥 2 + 5𝑥 + 3 = −4𝑥 2 − 5𝑥 − 3 (𝑝𝑡𝑣𝑛)
𝑇𝐻2: 𝑥 + 𝑦 ≠ 0 𝑡𝑎 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ℎ𝑜 𝑥 + 𝑦 𝑐ủ𝑎 𝑝𝑡(1):
1
𝑥𝑦
1
𝑥𝑦
√ −

+

=1
2 (𝑥 + 𝑦)2
3 3(𝑥 + 𝑦)2

đặ𝑡 𝑎 =

𝑥𝑦
1
1−𝑎


𝑡𝑎
đượ𝑐
𝑝𝑡
𝑛ℎư
𝑠𝑎𝑢:

𝑎
+
=1
(𝑥 + 𝑦)2
2
3
1
1 𝑎
1 4𝑎
√4 ( − 𝑎) ( − ) = +
2
3 3
6 3
1 𝑎 𝑎 𝑎2
1 4𝑎 16𝑎2
4( − − + ) =
+
+
6 6 3 3
36 9
9


1
23
𝑎 = (𝑛ℎậ𝑛 )ℎ𝑎𝑦 𝑎 = −
(𝑛ℎậ𝑛 )
4
4
đế𝑛 đâ𝑦 𝑡ℎì 𝑥é𝑡 𝑡ừ𝑛𝑔 𝑐á𝑖 ^^!
(4𝑥 2 + 1)𝑥 + (𝑦 − 3)√5 − 2𝑦 = 0(1)
𝑏à𝑖 24: {
4𝑥 2 + 𝑦 2 + 2√3 − 4𝑥 = 7(2)
𝑔𝑖ả𝑖
5
2
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: {
3
𝑥≤
4
𝑦≤

để ý 𝑝𝑡(1): (8𝑥 2 + 2)𝑥 + (2𝑦 − 6)√5 − 2𝑦 = 0
3

(2𝑥)3 + 2𝑥 = (√5 − 2𝑦) + (√5 − 2𝑦)
𝑥é𝑡 𝑓(𝑡) = 𝑡 3 + 𝑡 𝑣ậ𝑦 𝑓 ′ (𝑡) = 3𝑡 2 + 1 > 0
𝑡ứ𝑐 𝑓(𝑡) đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓(2𝑥) = 𝑓(√5 − 2𝑦)
2𝑥 = √5 − 2𝑦 (𝑡ừ đâ𝑦 𝑡ℎế 𝑣à𝑜 (2))
1
𝑥 2 + 𝑦 2 = (1)
5
𝑏à𝑖 25: {
57
4𝑥 2 + 3𝑥 −
= −𝑦(3𝑥 + 1)(2)
25
𝑔𝑖ả𝑖
25𝑥 2 + 25𝑦 2 = 5(1)
{
200𝑥 2 + 150𝑥 − 114 = −150𝑥𝑦 − 50𝑦(2)
𝑙ấ𝑦 𝑝𝑡(1) + 𝑝𝑡(2): (15𝑥)2 + (5𝑦)2 + 52 + 2(15𝑥)(5𝑦) + 2(5𝑦)5 + 2(15𝑥). 5
= 144
(15𝑥 + 5𝑦 + 5)2 = 122


[

15𝑥 + 5𝑦 + 5 = 12
15𝑥 + 5𝑦 + 5 = −12
4

(𝑥 4 + 𝑦). 3𝑦−𝑥 = 1(1)

𝑏à𝑖 26: {
4
8(𝑥 4 + 𝑦) − 6𝑥 −𝑦 = 0(2)
𝑔𝑖ả𝑖
4

6𝑥 −𝑦
4
𝑙ấ𝑦 (2)𝑡ℎế (1): 𝑥 + 𝑦 =
8
4

6𝑥 −𝑦 𝑦−𝑥 4
4
.3
= 1 => 2𝑥 −𝑦 = 23 => 𝑥 4 − 𝑦 = 3(∗)
8
𝑙ấ𝑦 (∗)𝑡ℎế (2): 8(3 + 2𝑦) − 63 = 0 𝑣ậ𝑦 𝑦 = 12
𝑥 2 + 𝑦 2 − 𝑦 = (2𝑥 + 1)(𝑦 − 1)(1)
5
𝑏à𝑖 27: {
(2)
√3𝑥 − 8 − √𝑦 =
𝑥 + 𝑦 − 12
𝑔𝑖ả𝑖
𝑦≥0
8
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: {
𝑥≥
3
𝑝𝑡(1): 𝑥 2 + 𝑦 2 − 𝑦 = 2𝑥𝑦 − 2𝑥 + 𝑦 − 1
(𝑥 − 𝑦)2 − 2(𝑥 − 𝑦) + 1 = 0
𝑥 − 𝑦 = −1 (𝑡ℎế 𝑣à𝑜 (2))
(𝑥𝑦)2 − 5𝑥 − 5𝑦 + 1 = 0(1)

𝑏à𝑖 28: {
(𝑥 + 𝑦 − 1)√𝑦 − 1 = (𝑦 − 2)√𝑥 + 𝑦(2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ {

𝑥+𝑦 ≥0
𝑦≥1


𝑝𝑡(2): (𝑥 + 𝑦)√𝑦 − 1 − √𝑦 − 1 = (𝑦 − 1)√𝑥 + 𝑦 − √𝑥 + 𝑦
√(𝑥 + 𝑦)(𝑦 − 1)(√𝑥 + 𝑦 − √𝑦 − 1) = √𝑦 − 1 − √𝑥 + 𝑦
𝑣ậ𝑦 𝑥 = −1 ℎ𝑎𝑦 √(𝑥 + 𝑦)(𝑦 − 1) = −1(𝑝𝑡𝑣𝑛)
𝑥 3 + 12𝑦 2 + 𝑥 + 2 = 8𝑦 3 + 8𝑦(1)

𝑏à𝑖 29: { 2
3
2𝑥 + (2𝑦 − 1)2 = √𝑥 3 + 8𝑦 + 2(2)
𝑔𝑖ả𝑖
𝑝𝑡(1): 𝑥 3 + 𝑥 + 2 = 8𝑦 3 + 8𝑦 − 12𝑦 2
𝑥 3 + 𝑥 = (2𝑦 − 1)3 + (2𝑦 − 1)
𝑥é𝑡 𝑓(𝑡) = 𝑡 3 + 𝑡 𝑣ậ𝑦 𝑓 ′ (𝑡) = 3𝑡 2 + 1 > 0
𝑛ê𝑛 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑣ậ𝑦 𝑓(𝑥) = 𝑓(2𝑦 − 1)
𝑣ậ𝑦 𝑥 = 2𝑦 − 1 (∗)𝑡ℎế (∗)𝑣à𝑜 (2)
2 + √𝑥 2 𝑦 4 + 2𝑥𝑦 2 − 5𝑦 4 + 1 = 2𝑦 2 (4 − 𝑥)(1)
𝑏à𝑖 30: {
2𝑥 + √𝑥 − 𝑦 2 = 5(2)
𝑔𝑖ả𝑖
𝑥 − 𝑦2 ≥ 0
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ {
(𝑥𝑦 2 + 1)2 − 5𝑦 4 ≥ 0
𝑝𝑡(1): 2 + √(𝑥𝑦 2 + 1)2 − 5𝑦 4 = 8𝑦 2 − 2𝑥𝑦 2
2(1 + 𝑥𝑦 2 ) + √(𝑥𝑦 2 + 1)2 − 5𝑦 4 = 8𝑦 2
𝑇𝐻1: 𝑦 2 = 0
𝑇𝐻2: 𝑦 2 ≠ 0 𝑡𝑎 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ℎ𝑜 𝑦 2 𝑐ủ𝑎 𝑝𝑡(1):


(𝑥𝑦 2 + 1)2
2(1 + 𝑥𝑦 2 )

+
−5=8
𝑦2
𝑦4
𝑥𝑦 2 + 1
đặ𝑡 𝑎 =
𝑝𝑡 𝑡𝑟ê𝑛 𝑡ℎà𝑛ℎ:
𝑦2
𝑎2 − 5 = 64 − 32𝑎 + 4𝑎2
𝑎 = 3 𝑣à 𝑎 =
𝑥+

𝑦

22
3
+ 𝑦 2 = 0(1)

√𝑥 2 + 1 + 𝑥
𝑏à𝑖 31: {
𝑥 2 + 2𝑦 2 √𝑥 2 + 1 + 𝑦 4 = 3𝑦 2 (2)
𝑔𝑖ả𝑖
𝑝𝑡(1): 𝑥 + 𝑦 2 + 𝑦 (√𝑥 2 + 1 − 𝑥) = 0
√𝑥 2

−(𝑥 + 𝑦 2 )
+1−𝑥 =
𝑦

𝑝𝑡(2): (𝑥 + 𝑦 2 )2 − 2𝑥𝑦 2 + 2𝑦 2 √𝑥 2 + 1 = 3𝑦 2
(𝑥 + 𝑦 2 )2 + 2𝑦 2 (√𝑥 2 + 1 − 𝑥) = 3𝑦 2
𝑙ấ𝑦 (1)𝑡ℎế (2): (𝑥 + 𝑦 2 )2 − 2(𝑥 + 𝑦 2 )𝑦 − 3𝑦 2 = 0
𝑥 + 𝑦 2 = 3 ℎ𝑎𝑦 𝑥 + 𝑦 2 = −1(𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) 𝑔𝑖ả𝑖 𝑏ì𝑛ℎ 𝑡ℎườ𝑛𝑔)
𝑥(𝑥𝑦 + 𝑥)2 + (𝑥 + 1)2 = 𝑥 3 (𝑦 2 + 𝑦 + 1) + 𝑥 2 (𝑦 − 1) + 5𝑥(1)
𝑏à𝑖 32: {
4𝑥 3 𝑦 + 7𝑥 2 + 2𝑥 2 √𝑦 + 1 = 2𝑥 + 1(2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 𝑦 ≥ −1
𝑝𝑡(1): 𝑥 3 𝑦 2 + 2𝑥 3 𝑦 + 𝑥 3 + 𝑥 2 + 2𝑥 + 1 = 𝑥 3 𝑦 2 + 𝑥 3 𝑦 + 𝑥 3 + 𝑥 2 𝑦 − 1 + 5𝑥


𝑥 3 𝑦 − 𝑥 2 𝑦 + 𝑥 2 − 3𝑥 + 2 = 0
𝑥 2 𝑦(𝑥 − 1) + (𝑥 − 1)(2𝑥 − 1) = 0
𝑣ậ𝑦 𝑥 = 1 ℎ𝑎𝑦 𝑥 2 𝑦 + 2𝑥 − 1
= 0 (đế𝑛 đâ𝑦 𝑡ℎì 𝑚ấ𝑦 𝑏ạ𝑛 𝑔𝑖ả𝑖 𝑏ì𝑛ℎ 𝑡ℎườ𝑛𝑔 đượ𝑐 𝑟ò𝑖 ‼!)
𝑥 4 + 2𝑥 3 − 5𝑥 2 + 𝑦 2 − 6𝑥 − 11 = 0(1)
𝑏𝑎𝑖 33: {

2

𝑥 +𝑥 =

3√𝑦 2 − 7 − 6
√𝑦 2 − 7

(2)

𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: 𝑦 2 − 7 > 0
𝑝𝑡(1): 𝑥 4 + 2𝑥 3 + 𝑥 2 − 6𝑥 2 − 6𝑥 + 𝑦 2 − 11 = 0
(𝑥 2 + 𝑥)2 − 6(𝑥 2 + 𝑥) + 𝑦 2 − 7 − 4 = 0
đặ𝑡 𝑎 = 𝑥 2 + 𝑥, 𝑏 = √𝑦 2 − 7
2
2
{𝑎 − 6𝑎 + 𝑏 − 4 = 0
𝑎𝑏 = 3𝑏 − 6

𝑡ℎế (2)𝑣à𝑜 (1): 𝑎 = 3 −

6
𝑏

6 2
6
(3 − ) − 6 (3 − ) + 𝑏 2 − 4 = 0
𝑏
𝑏
9−

36 36
36
+ 2 − 18 +
+ 𝑏2 − 4 = 0
𝑏
𝑏
𝑏
36
+ 𝑏 2 − 13 = 0
𝑏2
𝑏 2 = 3 𝑣à 𝑏2 = 9
𝑛ℎậ𝑛 𝑏 = √3 𝑣à 𝑏 = 3


𝑣ớ𝑖 𝑦 2 − 7 = √3 𝑡ℎì 𝑦 = ±√√3 + 7
𝑣ớ𝑖 𝑦 2 − 7 = 3 𝑡ℎì 𝑦 = ±√10
4𝑥 2 = (√𝑥 2 + 1 + 1) (𝑥 2 − 𝑦 3 + 3𝑦 − 2)(1)
𝑏à𝑖 34: {
1 − 𝑥2
2
2
𝑥 + (𝑦 + 1) = 2 (1 +
) (2)
𝑦
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 𝑦 ≠ 0
𝑦 + 1 − 𝑥2
𝑝𝑡(2): 𝑥 + 𝑦 + 2𝑦 + 1 = 2 (
)
𝑦
2

2

𝑥 2 𝑦 + 𝑦 3 + 2𝑦 2 + 𝑦 = 2𝑦 + 2 − 2𝑥 2
𝑥 2 (𝑦 + 2) + 𝑦 2 (𝑦 + 2) − (𝑦 + 2) = 0
𝑣ậ𝑦 𝑦 = −2 ℎ𝑎𝑦 𝑥 2 + 𝑦 2 = 1
(∗)𝑣ớ𝑖 𝑦 = −2 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1):
4𝑥 2 = (√𝑥 2 + 1 + 1) 𝑥 2
𝑣ậ𝑦 𝑥 = 0 ℎ𝑎𝑦 𝑥 = ±2√2
(∗∗)𝑣ớ𝑖 𝑥 2 = 1 − 𝑦 2 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2):
4(1 − 𝑦 2 ) = (√2 − 𝑦 2 + 1) (−𝑦 3 − 𝑦 2 + 3𝑦 − 1)(𝑙𝑖ê𝑛 ℎợ𝑝 𝑣𝑠 𝑛𝑔ℎ𝑖ệ𝑚 𝑦 = 1)
𝑥 3 + 3𝑥𝑦 2 + 3𝑥 = 𝑦(3𝑥 2 + 𝑦 2 + 3)(1)
𝑏à𝑖 35: { 4 2
3𝑥 (𝑦 − 2) − 2𝑥 3 (𝑦 2 + 8) + 2𝑥(15𝑦 − 6) = 0(2)
𝑔𝑖ả𝑖
𝑝𝑡(1): 𝑥 3 + 3𝑥𝑦 2 + 3𝑥 = 3𝑥 2 𝑦 + 𝑦 3 + 3𝑦


(𝑥 − 𝑦)(𝑥 2 + 𝑥𝑦 + 𝑦 2 ) − 3𝑥𝑦(𝑥 − 𝑦) + 3(𝑥 − 𝑦) = 0
𝑣ậ𝑦 𝑥 = 𝑦 ℎ𝑎𝑦 (𝑥 − 𝑦)2 + 3 = 0(𝑝𝑡𝑣𝑛 𝑑𝑜 (𝑥 − 𝑦)2 + 3 > 0)
𝑡ℎế 𝑥 = 𝑦 𝑣à𝑜 𝑝𝑡(2):
3𝑦 6 − 6𝑦 4 − 2𝑦 5 − 16𝑦 3 + 30𝑦 2 − 12𝑦 = 0
𝑦 = 0 ℎ𝑎𝑦 3𝑦 5 − 6𝑦 3 − 2𝑦 4 − 16𝑦 2 + 30𝑦 − 12 = 0
𝑣ậ𝑦 𝑥 = 𝑦 = 0 ℎ𝑎𝑦 𝑥 = 𝑦 = 2
√7𝑥 + 𝑦 − √2𝑥 + 𝑦 = 4(1)
𝑏à𝑖 36: {
2√2𝑥 + 𝑦 − √5𝑥 + 8 = 2(2)
𝑔𝑖ả𝑖
7𝑥 + 𝑦 ≥ 0
8
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: { 𝑥 ≥ −
5
2𝑥 + 𝑦 ≥ 0
đặ𝑡 ∶ 𝑎2 = 7𝑥 + 𝑦, 𝑏 2 = 2𝑥 + 𝑦 (đ𝑘 𝑎, 𝑏 ≥ 0)
𝑣ậ𝑦 𝑎2 − 𝑏 2 = 5𝑥
{

𝑎 − 𝑏 = 4(1)
2𝑏 − √𝑎2 − 𝑏 2 + 8 = 2(2)

𝑙ấ𝑦 (1)𝑡ℎế (2): 2𝑏 − √(4 + 𝑏)2 − 𝑏 2 + 8 = 2
24 + 8𝑏 = 4𝑏 2 − 8𝑏 + 4
𝑏 = 5(𝑛ℎậ𝑛) ℎ𝑎𝑦 𝑏 = −1 (𝑙𝑜ạ𝑖)
(∗)𝑣ớ𝑖 𝑏 = 5 𝑡ℎì 𝑎 = 9 (𝑡ℎõ𝑎)
56
13
2𝑥 + 𝑦 = 25
=> 𝑥 =
,𝑦 =
{
7𝑥 + 𝑦 = 81
5
5


3

3
𝑥 + √𝑥 2 − 2𝑥 + 5 = 3𝑦 + √𝑦 2 + 4(1)
𝑏à𝑖 37: {
(𝑥 − 𝑦)(𝑥 + 𝑦 − 3) = −1(2)

𝑔𝑖ả𝑖
3

3

𝑝𝑡(1): 𝑥 + √(𝑥 − 1)2 + 4 = 3𝑦 + √𝑦 2 + 4
𝑝𝑡(2): 𝑥 2 − 𝑦 2 − 3𝑥 + 3𝑦 + 1 = 0
𝑙ấ𝑦 𝑝𝑡(2)𝑡ℎế 𝑝𝑡(1): 3𝑦 = 𝑦 2 − 𝑥 2 + 3𝑥 − 1
3

3

𝑥 + √(𝑥 − 1)2 + 4 = 𝑦 2 − 𝑥 2 + 3𝑥 − 1 + √𝑦 2 + 4
3

3

(𝑥 − 1)2 + √(𝑥 − 1)2 + 4 = 𝑦 2 + √𝑦 2 + 4
3

𝑥é𝑡 𝑓(𝑡) = 𝑡 + √𝑡 + 4
𝑑ễ 𝑡ℎấ𝑦 𝑓 ′ (𝑡) > 0 𝑛ê𝑛 𝑓(𝑡) đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑣ậ𝑦 𝑓((𝑥 − 1)2 ) = 𝑓(𝑦 2 )
𝑣ậ𝑦 𝑥 − 1 = 𝑦 ℎ𝑎𝑦 𝑥 − 1 = −𝑦
𝑥 3 − 𝑥 2 + 𝑥 = 𝑦√𝑦 − 1 − 𝑦 + 1(1)
𝑏à𝑖 38: {
𝑥 3 + 4𝑥 2 + 1 = 𝑦 2 (2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑦 ≥ 1
3

2

𝑝𝑡(1): 𝑥 3 − 𝑥 2 + 𝑥 = (√𝑦 − 1) − (√𝑦 − 1) + (√𝑦 − 1)
𝑥é𝑡 𝑓(𝑡) = 𝑡 3 − 𝑡 2 + 𝑡
𝑓

′ (𝑡)

2

= 3𝑡 − 2𝑡 + 1 = (√3𝑡 −

1
√3

2

) +

2
>0
3

𝑣ậ𝑦 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓(𝑥) = 𝑓(√𝑦 − 1) 𝑣ậ𝑦 𝑥 2 = 𝑦 − 1(đ𝑘 𝑥 ≥ 0)


3

𝑏à𝑖 39: {

√𝑥 3 + 2 + 𝑥 2 (𝑥 3 + 𝑥 − 𝑦 − 1) = 3√𝑦 + 3 + 𝑦 + 1(1)
3

√4𝑥 2 − 2 − 4𝑥 2 − 25𝑥 − 24 = 2𝑦(2)
𝑔𝑖ả𝑖

3

𝑝𝑡(1): √𝑥 3 + 2 − 3√𝑦 + 3 + 𝑥 3 (𝑥 2 + 1) − 𝑥 2 (𝑦 + 1) − (𝑦 + 1) = 0
𝑥3 − 𝑦 − 1
3

2

3

3
+ 3)
(√𝑥 3 + 2) + √(𝑥 3 + 2)(𝑦 + 3) + (√𝑦

2

+ (𝑥 3 − 𝑦 − 1)(𝑥 2 + 1) = 0

𝑣ậ𝑦 𝑥 3 − 𝑦 − 1
= 0 ℎ𝑎𝑦

1
2

3

3

3

2

+ (𝑥 2 + 1)

(√𝑥 3 + 2) + √(𝑥 3 + 2)(𝑦 + 3) + (√𝑦 + 3)
1
= 0 (𝑝𝑡𝑣𝑛 𝑑𝑜
2
2
3
3
3
+ 3)
(√𝑥 3 + 2) + √(𝑥 3 + 2)(𝑦 + 3) + (√𝑦
+ (𝑥 2 + 1) > 0 )
𝑣ậ𝑦 𝑥 3 − 𝑦 − 1 = 0 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)
3

5√(𝑥 + 7)2 3𝑥 2 + 21𝑥
𝑦 −
+
= 0(1)
𝑦
2𝑦
𝑏à𝑖 40:
3𝑥 2 + 21𝑥
3
− 2(𝑦 + 5)√𝑥 + 7 + 4𝑦 2 = 0(2)
3
{ √𝑥 + 7
2

𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≠ −7 𝑣à 𝑦 ≠ 0
3

𝑝𝑡(1): 3𝑥 2 + 21𝑥 = 10√(𝑥 + 7)2 − 2𝑦 3
3

3

3

𝑙ấ𝑦 𝑝𝑡(1) 𝑡ℎế 𝑝𝑡(2): 10√(𝑥 + 7)2 − 2𝑦 3 − 2(𝑦 + 5)√(𝑥 + 7)2 + 4𝑦 2 √𝑥 + 7
=0
3

3

−2𝑦 3 − 2𝑦 √(𝑥 + 7)2 + 4𝑦 2 √𝑥 + 7 = 0


2

3

3

𝑦 = 0 ℎ𝑎𝑦 − 2𝑦 2 − 2(√𝑥 + 7) + 4𝑦 √𝑥 + 7 = 0
𝑦 = 0(𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑦 3 = 𝑥 + 7 (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2))
𝑏à𝑖 41: {

𝑦 − 3𝑥 − 3 + 4√𝑥 − 1 = 0(1)
√𝑦 − 1 + 𝑦(𝑦 2 + 𝑦 − 1) = 𝑥 2 (𝑦 + 1)2 + 𝑥 + 1(2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 𝑥 ≥ 1, 𝑦 ≥ 1

𝑝𝑡(2): √𝑦 − 1 + 𝑦 3 + 𝑦 2 − 𝑦 = (𝑥𝑦 + 𝑥)2 + 𝑥 + 1
√𝑦 − 1 + 𝑦 2 (𝑦 + 1) − (𝑦 + 1) = (𝑥𝑦 + 𝑥)2 + 𝑥
√𝑦 − 1 + (𝑦 + 1)2 (𝑦 − 1) = 𝑥 2 (𝑦 + 1)2 + 𝑥
𝑦 − 1 − 𝑥2
√𝑦 − 1 + 𝑥

= (𝑦 + 1)2 (𝑥 2 − 𝑦 + 1)(𝑥 2 + 𝑦 − 1)

𝑣ậ𝑦 𝑥 2 − 𝑦 + 1 = 0 ℎ𝑎𝑦 −

1

√𝑦 − 1 + 𝑥
< 0 𝑚à 𝑉𝑃 > 0)

= (𝑦 + 1)2 (𝑥 2 + 𝑦 − 1)(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇

(∗)𝑣ớ𝑖 𝑥 2 − 𝑦 + 1 = 0 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) 𝑡𝑎 𝑡ì𝑚 đượ𝑐 𝑥, 𝑦
𝑥 √𝑥(𝑦 2 − 2𝑦 + 2) = (𝑦 3 − 3𝑦 2 + 3𝑦 − 1)(𝑥 + 1)(1)
𝑏à𝑖 42: {
2𝑥 2 − 11𝑦 2 + 22𝑦 − 2√2𝑥 − 1 = 0(2)
𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≥

1
2

3

2

𝑝𝑡(1): (√𝑥) ((𝑦 − 1)2 + 1) = (𝑦 − 1)3 ((√𝑥) + 1)
3

3

2

(√𝑥) (𝑦 − 1)2 + (√𝑥) = (𝑦 − 1)3 (√𝑥) + (𝑦 − 1)3


𝑥(𝑦 − 1)2 (√𝑥 − 𝑦 + 1) + (√𝑥 − 𝑦 + 1)(𝑥 + √𝑥(𝑦 − 1) + (𝑦 − 1)2 ) = 0
𝑣ậ𝑦 √𝑥 − 𝑦 + 1 = 0 ℎ𝑎𝑦 𝑥(𝑦 − 1)2 + 𝑥 + √𝑥(𝑦 − 1) + (𝑦 − 1)2
= 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 > 0)
(∗)𝑣ớ𝑖 𝑦 = √𝑥 + 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)𝑡𝑎 đượ𝑐 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = 1 𝑣à 𝑥 = 5
3

𝑏à𝑖 43: {

√𝑥 2 − 1 + 𝑥 3 = 3√𝑦 − 1 + 𝑥𝑦(1)

(𝑥, 𝑦 ∈ 𝑅), 𝑥 ≥ 0
16𝑦 − 24𝑦 + 8√3 − 4𝑥 − 3 = 0(2)
2

𝑔𝑖ả𝑖
đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 0 ≤ 𝑥 ≤
𝑝𝑡(1):

3
4

𝑥2 − 𝑦
2

3

3

3

(√𝑥 2 − 1) + √(𝑥 2 − 1)(𝑦 − 1) + (√𝑦 − 1)
𝑣ậ𝑦 𝑥 2 = 𝑦 ℎ𝑎𝑦

2

= −𝑥(𝑥 2 − 𝑦)

1
2

3

3

3
− 1)
(√𝑥 2 − 1) + √(𝑥 2 − 1)(𝑦 − 1) + (√𝑦
= −𝑥(𝑝𝑡𝑣𝑛 𝑑𝑜 đ𝑘)

1
(∗)𝑣ớ𝑖 𝑥 2 = 𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)𝑡𝑎 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = (𝑡ℎõ𝑎)
2
3

𝑦 2 − √3𝑥 + 2 = 0(1)

𝑏à𝑖 44: {
3
𝑥 + 2√𝑥 − 1 = 𝑦 6 + 3𝑦 4 + 5𝑦 2 + 4(2)
𝑔𝑖ả𝑖
3

𝑝𝑡(2): 𝑥 + 2√𝑥 − 1 = 𝑦 6 + 3𝑦 4 + 3𝑦 2 + 2𝑦 2 + 4
3

3

3

(√𝑥 − 1) + 2√𝑥 − 1 = (𝑦 2 + 1)3 + 2(𝑦 2 + 1)
𝑥é𝑡 𝑓(𝑡) = 𝑡 3 + 2𝑡 𝑣ậ𝑦 𝑓 ′ (𝑡) = 3𝑡 2 + 2 > 0

2


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