các dạng bài phương trình và hệ phương trình nâng cao hay nhất

𝑐á𝑐 𝑏à𝑖 ℎệ phương trình ℎ𝑎𝑦 𝑣à 𝑘ℎó (Nguyễn Trường Phát TP.HCM)

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

2𝑡ℎế 𝑥 = 3 𝑣à 𝑦 = 16 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎

𝑡ℎế 𝑥 = 3 𝑣à 𝑦 = −3

2 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑐ũ𝑛𝑔 𝑡ℎõ𝑎 (∗)𝑣ớ𝑖 √𝑥 − 2𝑦 + 1 = 0 𝑡ℎế 𝑣à𝑜 (2)

√𝑥 + √𝑥 + 2 + √𝑥 − √𝑥 + 13 = 3(𝑙𝑖ê𝑛 ℎợ𝑝 𝑣ớ𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = 1)

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

2)

𝑏à𝑖 2: {2𝑥 + 1 = 𝑦

2+ 4𝑦√𝑥(1)

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 = 𝑦2 − 2𝑦 + 1

√𝑥 = 0 ℎ𝑎𝑦 2√𝑥 = 𝑦2+ 2𝑦 + 1 𝑣ớ𝑖 𝑥 = 0 𝑡ℎì 𝑦 = ±1 𝑡ℎế 𝑣à𝑜 (2)𝑡𝑎 𝑛ℎậ𝑛 𝑥 = 0, 𝑦 = −1

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

(𝑥 + 𝑦)(2𝑥 − 𝑦) + 6𝑥 + 3𝑦 + 4 = 0(2)

𝑔𝑖ả𝑖

đ𝑘 để ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ {𝑥 ≥

13

𝑦 ≥ 1𝑝𝑡(2): 2𝑥2− 𝑥𝑦 + 2𝑥𝑦 − 𝑦2+ 6𝑥 + 3𝑦 + 4 = 0

2𝑥2+ 𝑥𝑦 − 𝑦2+ 6𝑥 + 3𝑦 + 4 = 0 2𝑥2+ 𝑥(𝑦 + 6) − 𝑦2+ 3𝑦 + 4 = 0 𝑑𝑒𝑛𝑡𝑎 = 𝑦2 + 12𝑦 + 36 − 8(−𝑦2 + 3𝑦 + 4) = 9𝑦2− 12𝑦 + 4 = (3𝑦 − 2)2

𝑥4𝑦 + 𝑥4+ 4𝑥2𝑦 + 4𝑥2+ 4𝑦 + 4 = 𝑦2𝑥2+ 𝑦2 + 4𝑦𝑥2+ 4𝑦 + 4𝑥2+ 4

𝑥4𝑦 + 𝑥4 = 𝑥2𝑦2+ 𝑦2

𝑥2𝑦(𝑥2− 𝑦) = (𝑦 − 𝑥2)(𝑦 + 𝑥2)

𝑣ậ𝑦 𝑥2 = 𝑦 ℎ𝑎𝑦 𝑥2𝑦 = −(𝑦 + 𝑥2)

𝑣ớ𝑖 𝑥2 = 𝑦 𝑡ℎế 𝑣à𝑜 (2): −3𝑦2− 5𝑦 + 8 = 0[ 𝑦 = 1

𝑦 = −83𝑣ớ𝑖 𝑥2𝑦 + 𝑦 = −𝑥2 => 𝑦 = − 𝑥

𝑥é𝑡 𝑓(𝑡) = 𝑡2+ 𝑡 + √𝑡 + 1 𝑣ậ𝑦 𝑓′(𝑡) = 2𝑡 + 1 + 1

2√𝑡 + 1 > 0 𝑡ứ𝑐 𝑓(𝑥 + 1) = 𝑓(2𝑦)𝑣ậ𝑦 𝑥 + 1 = 2𝑦(3)

𝑙ấ𝑦 (3)𝑡ℎế (2): (2𝑦 − 1)2+ 2𝑦2− 2(2𝑦 − 1) + 𝑦 − 2 = 0

4𝑦2− 4𝑦 + 1 + 2𝑦2− 4𝑦 + 2 + 𝑦 − 2 = 0

6𝑦2− 7𝑦 + 1 = 0 [𝑦 = 1

𝑦 =16𝑣ớ𝑖 𝑦 = 1 𝑣ậ𝑦 𝑥 = 1

𝑣ớ𝑖 𝑦 = 1

6 𝑣ậ𝑦 𝑥 = −

23

𝑏à𝑖 6: { 𝑥

3𝑦√𝑥 + 𝑥3𝑦2 = 2𝑥4√𝑥 + 2𝑥4𝑦(1)𝑦√𝑥 (√2𝑥2− 6 − 1) = √5(2𝑥2− 6)(2)

𝑔𝑖ả𝑖 𝑝𝑡(1): 𝑥3𝑦(√𝑥 + 𝑦) = 2𝑥4(√𝑥 + 𝑦) 𝑣ậ𝑦 √𝑥 = −𝑦 ℎ𝑎𝑦 𝑥 = 0 ℎ𝑎𝑦 𝑦 = 2𝑥

𝑏à𝑖 8: {2𝑥(𝑥

2+ 3) − 𝑦(𝑦2+ 3) = 3𝑥𝑦(𝑥 − 𝑦)(1)(𝑥2− 2)2 = 4(2 − 𝑦)(2)

𝑔𝑖ả𝑖 𝑝𝑡(1)2𝑥3+ 6𝑥 − 𝑦3− 3𝑦 = 3𝑥2𝑦 − 3𝑥𝑦22𝑥3− 3𝑥2𝑦 + 3𝑥𝑦2− 𝑦3+ 6𝑥 − 3𝑦 = 0 (2𝑥 − 𝑦)(𝑥2− 𝑥𝑦 + 𝑦2) + 3(2𝑥 − 𝑦) = 0 2𝑥 = 𝑦 ℎ𝑎𝑦 𝑥2− 𝑥𝑦 + 𝑦2 = −3 (𝑝𝑡 𝑣𝑛)

𝑣ờ𝑖 2𝑥 = 𝑦 𝑡ℎế 𝑣à𝑜 (2):

(𝑥2− 2)2 = 4(2 − 2𝑥)

𝑥4− 4𝑥2+ 4 = 8 − 8𝑥 (𝑥2+ 2𝑥 − 2)(… ) = 0 => 𝑥 = −1 ± √3

8𝑥3− 12𝑥2+ 6𝑥 − 1 + 𝑥 = (𝑦 −1

2) √𝑦 − 1 +1

2

𝑥2+ 𝑥𝑦2− 𝑦2 − 𝑦3 = 𝑥𝑦 + 𝑥 𝑥(𝑥 + 𝑦2) = 𝑥(𝑦 + 1) + 𝑦2(𝑦 + 1) 𝑥(𝑥 + 𝑦2) = (𝑥 + 𝑦2)(𝑦 + 1)

𝑣ậ𝑦 𝑥 + 𝑦2 = 0 ℎ𝑎𝑦 𝑥 = 𝑦 + 1 𝑣ớ𝑖 𝑥 + 𝑦2 = 0 (𝑝𝑡𝑣𝑛)𝑑𝑜 𝑥 + 𝑦2 > 0

𝑣ớ𝑖 𝑦 = 𝑥 − 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)

√𝑥3+ 4𝑥 = 1 +(𝑥 + 1)2

𝑥2+ 2𝑥 + 133√𝑥3+ 4𝑥 = 𝑥2+ 2𝑥 + 4(𝑙𝑖ê𝑛 ℎợ𝑝 𝑣ớ𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = 2)

𝑣ớ𝑖 𝑥 = 2 𝑡ℎì 𝑦 = 1 𝑣ậ𝑦 ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: (2,1)

3 = 19𝑥3(1)

𝑦 + 𝑥𝑦2 = −6𝑥2(2)𝑔𝑖ả𝑖

𝑇𝐻1: 𝑥 = 0 𝑡ℎế 𝑣à𝑜 (1) 𝑝𝑡𝑣𝑛 𝑇𝐻2: 𝑥 ≠ 0 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ủ𝑎 𝑝𝑡(1)𝑐ℎ𝑜 𝑥3:

1

𝑥3+ 𝑦3 = 19 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ủ𝑎 𝑝𝑡(2)𝑐ℎ𝑜 𝑥2:

𝑔𝑖ả𝑖 𝑡ℎế (2) 𝑣à𝑜 (1): 16𝑥3𝑦3− 9𝑦3 = 𝑦3(2𝑥 − 1)(4𝑥2+ 6𝑥 + 1) 𝑣ậ𝑦 𝑦 = 0 ℎ𝑎𝑦 16𝑥3− 9 = 8𝑥3+ 12𝑥2+ 2𝑥 − 4𝑥2− 6𝑥 − 1

đặ𝑡 𝑡ℎử ℎệ 𝑠ố ∶ 𝑥 − 2𝑦 + 𝑎𝑏 + (𝑎 + 𝑏)√𝑥 − 2𝑦 = 0

√−𝑦2− 𝑦 + 1 − 𝑦2+ 𝑦 + 1 = 0( 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑦 = −1) 𝑣ậ𝑦 𝑦 = −1 𝑡ℎì 𝑥 = −2 𝑡ℎế 𝑣à𝑜 đ𝑘 𝑡ℎấ𝑦 𝑡ℎõ𝑎

(∗∗)𝑣ớ𝑖 𝑥 = −𝑦2 − 𝑦 𝑡ℎế 𝑣à𝑜 (2):

√𝑦2− 𝑦 + 1 − 𝑦2+ 𝑦 + 1 = 0

3 − 3𝑥𝑦 + 3𝑥 = 6(1)

𝑥2√𝑦 − 1 − 2𝑥𝑦 = 1 − 2𝑥(2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: 𝑦 ≥ 1 𝑝𝑡(2): 6𝑥2√𝑦 − 1 − 12𝑥𝑦 + 12𝑥 = 6

𝑥3− 6𝑥2√𝑦 − 1 + 9𝑥𝑦 − 9𝑥 = 0

𝑥 = 0 ℎ𝑎𝑦 𝑥2− 6𝑥√𝑦 − 1 + 9(𝑦 − 1) = 0(𝑝𝑡𝑣𝑛)

(∗)𝑣ớ𝑖 𝑥 = 0 𝑡ℎế 𝑣à𝑜 (2)𝑡𝑎 𝑐ó 𝑝𝑡𝑣𝑛

(𝑥 − 3√𝑦 − 1)2 = 0 (∗)𝑣ớ𝑖 𝑥 = 3√𝑦 − 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2):

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ 𝑦 ≥ −1 𝑝𝑡(1): (𝑥3+ 3𝑥2+ 3𝑥 + 1) + (𝑥 + 1) = (√𝑦 + 1)3+ √𝑦 + 1

(𝑥 + 1)3+ (𝑥 + 1) = (√𝑦 + 1)3+ √𝑦 + 1 𝑥é𝑡 𝑓(𝑡) = 𝑡3+ 𝑡 𝑣ậ𝑦 𝑡ℎì 𝑓′(𝑡) = 3𝑡2+ 1 > 0

𝑡ứ𝑐 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓(𝑥 + 1) = 𝑓(√𝑦 + 1)

𝑑𝑒𝑛𝑡𝑎 = 𝑥2 − 8𝑥 + 16 − 4𝑥2+ 12𝑥 − 16 = 0 ≤ 𝑥 ≤ 4

3𝑝𝑡(2): 𝑥2+ 2𝑥 + 27𝑥2 ≤ (4

𝑣ậ𝑦 𝑑ấ𝑢 𝑏ằ𝑛𝑔 𝑥ả𝑦 𝑟𝑎 𝑥 = 4

3 𝑛ê𝑛 𝑦 =

43𝑏à𝑖 23: { 𝑥𝑦

2(√𝑥2+ 1 + 1) = 3√𝑦2+ 9 + 3𝑦(1)(3𝑥 − 1)√𝑥2𝑦 + 𝑥𝑦 − 5 − 4𝑥3+ 3𝑥3𝑦 − 7𝑥 = 0(2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 𝑥𝑦(𝑥 + 1) ≥ 5

𝑇𝐻1: 𝑥é𝑡 𝑦2 = 0 𝑇𝐻2: 𝑦2 ≠ 0 𝑛ê𝑛 𝑡𝑎 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ủ𝑎 𝑝𝑡(1) 𝑙à 𝑦2:

3+ 2𝑡2√𝑡4+ 𝑡2+ 1 > 0

𝑣ậ𝑦 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓(𝑥) = 𝑓 (3

𝑦) (𝑡ừ đâ𝑦 𝑡ℎế 𝑣à𝑜 (2))

𝑏à𝑖 22: { 𝑥𝑦(𝑥 + 𝑦) = 2(1)

𝑦3 + 𝑦 + 6 = 7𝑥3+ 𝑥(2)

𝑔𝑖ả𝑖 { 3𝑥

2𝑦 + 3𝑥𝑦2 = 67𝑥3+ 𝑥 − 𝑦3 − 𝑦 = 6

−7𝑥3+ 3𝑥2𝑦 + 3𝑥𝑦2+ 𝑦3− 𝑥 + 𝑦 = 0

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

√1

2−

𝑥𝑦(𝑥 + 𝑦)2+ √1

3−

𝑥𝑦3(𝑥 + 𝑦)2 = 1

𝑎 = 1

4(𝑛ℎậ𝑛 )ℎ𝑎𝑦 𝑎 = −23

4 (𝑛ℎậ𝑛 ) đế𝑛 đâ𝑦 𝑡ℎì 𝑥é𝑡 𝑡ừ𝑛𝑔 𝑐á𝑖 ^^!

𝑥 ≤ 34

để ý 𝑝𝑡(1): (8𝑥2 + 2)𝑥 + (2𝑦 − 6)√5 − 2𝑦 = 0 (2𝑥)3+ 2𝑥 = (√5 − 2𝑦)3+ (√5 − 2𝑦) 𝑥é𝑡 𝑓(𝑡) = 𝑡3 + 𝑡 𝑣ậ𝑦 𝑓′(𝑡) = 3𝑡2+ 1 > 0 𝑡ứ𝑐 𝑓(𝑡) đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓(2𝑥) = 𝑓(√5 − 2𝑦)

2𝑥 = √5 − 2𝑦 (𝑡ừ đâ𝑦 𝑡ℎế 𝑣à𝑜 (2))

𝑏à𝑖 25: {

𝑥2+ 𝑦2 = 1

5(1)4𝑥2+ 3𝑥 −57

25 = −𝑦(3𝑥 + 1)(2)𝑔𝑖ả𝑖

2+ 25𝑦2 = 5(1)200𝑥2+ 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 = 1215𝑥 + 5𝑦 + 5 = −12

𝑏à𝑖 26: { (𝑥4+ 𝑦) 3𝑦−𝑥4 = 1(1)

8(𝑥4+ 𝑦) − 6𝑥4−𝑦 = 0(2)

𝑔𝑖ả𝑖 𝑙ấ𝑦 (2)𝑡ℎế (1): 𝑥4+ 𝑦 = 6

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: {

𝑦 ≥ 0

𝑥 ≥ 83𝑝𝑡(1): 𝑥2+ 𝑦2− 𝑦 = 2𝑥𝑦 − 2𝑥 + 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)

𝑏à𝑖 29: { 𝑥

3+ 12𝑦2+ 𝑥 + 2 = 8𝑦3 + 8𝑦(1)2𝑥2+ (2𝑦 − 1)2 = √𝑥3 3+ 8𝑦 + 2(2)

𝑔𝑖ả𝑖 𝑝𝑡(1): 𝑥3+ 𝑥 + 2 = 8𝑦3+ 8𝑦 − 12𝑦2

𝑥3+ 𝑥 = (2𝑦 − 1)3+ (2𝑦 − 1) 𝑥é𝑡 𝑓(𝑡) = 𝑡3 + 𝑡 𝑣ậ𝑦 𝑓′(𝑡) = 3𝑡2+ 1 > 0

𝑛ê𝑛 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑣ậ𝑦 𝑓(𝑥) = 𝑓(2𝑦 − 1)

𝑣ậ𝑦 𝑥 = 2𝑦 − 1 (∗)𝑡ℎế (∗)𝑣à𝑜 (2)

𝑏à𝑖 30: {2 + √𝑥

2𝑦4+ 2𝑥𝑦2 − 5𝑦4 + 1 = 2𝑦2(4 − 𝑥)(1)2𝑥 + √𝑥 − 𝑦2 = 5(2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ { 𝑥 − 𝑦

2 ≥ 0(𝑥𝑦2+ 1)2− 5𝑦4 ≥ 0𝑝𝑡(1): 2 + √(𝑥𝑦2 + 1)2− 5𝑦4 = 8𝑦2− 2𝑥𝑦2

2(1 + 𝑥𝑦2) + √(𝑥𝑦2+ 1)2− 5𝑦4 = 8𝑦2

𝑇𝐻1: 𝑦2 = 0 𝑇𝐻2: 𝑦2 ≠ 0 𝑡𝑎 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ℎ𝑜 𝑦2 𝑐ủ𝑎 𝑝𝑡(1):

2(1 + 𝑥𝑦2)

𝑦2 + √(𝑥𝑦2+ 1)2

𝑦4 − 5 = 8 đặ𝑡 𝑎 = 𝑥𝑦

√𝑥2+ 1 − 𝑥 = −(𝑥 + 𝑦

2)𝑦𝑝𝑡(2): (𝑥 + 𝑦2)2− 2𝑥𝑦2+ 2𝑦2√𝑥2+ 1 = 3𝑦2(𝑥 + 𝑦2)2+ 2𝑦2(√𝑥2+ 1 − 𝑥) = 3𝑦2𝑙ấ𝑦 (1)𝑡ℎế (2): (𝑥 + 𝑦2)2− 2(𝑥 + 𝑦2)𝑦 − 3𝑦2 = 0

𝑥 + 𝑦2 = 3 ℎ𝑎𝑦 𝑥 + 𝑦2 = −1(𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) 𝑔𝑖ả𝑖 𝑏ì𝑛ℎ 𝑡ℎườ𝑛𝑔)

𝑏à𝑖 32: {𝑥(𝑥𝑦 + 𝑥)

2+ (𝑥 + 1)2 = 𝑥3(𝑦2+ 𝑦 + 1) + 𝑥2(𝑦 − 1) + 5𝑥(1)4𝑥3𝑦 + 7𝑥2+ 2𝑥2√𝑦 + 1 = 2𝑥 + 1(2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 𝑦 ≥ −1 𝑝𝑡(1): 𝑥3𝑦2 + 2𝑥3𝑦 + 𝑥3+ 𝑥2+ 2𝑥 + 1 = 𝑥3𝑦2+ 𝑥3𝑦 + 𝑥3+ 𝑥2𝑦 − 1 + 5𝑥

𝑥3𝑦 − 𝑥2𝑦 + 𝑥2 − 3𝑥 + 2 = 0

𝑥2𝑦(𝑥 − 1) + (𝑥 − 1)(2𝑥 − 1) = 0 𝑣ậ𝑦 𝑥 = 1 ℎ𝑎𝑦 𝑥2𝑦 + 2𝑥 − 1

= 0 (đế𝑛 đâ𝑦 𝑡ℎì 𝑚ấ𝑦 𝑏ạ𝑛 𝑔𝑖ả𝑖 𝑏ì𝑛ℎ 𝑡ℎườ𝑛𝑔 đượ𝑐 𝑟ò𝑖 ‼!)

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: 𝑦2− 7 > 0 𝑝𝑡(1): 𝑥4 + 2𝑥3+ 𝑥2− 6𝑥2− 6𝑥 + 𝑦2− 11 = 0

𝑏2 + 𝑏2− 13 = 0

𝑏2 = 3 𝑣à 𝑏2 = 9 𝑛ℎậ𝑛 𝑏 = √3 𝑣à 𝑏 = 3

𝑣ớ𝑖 𝑦2− 7 = √3 𝑡ℎì 𝑦 = ±√√3 + 7 𝑣ớ𝑖 𝑦2− 7 = 3 𝑡ℎì 𝑦 = ±√10

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 𝑦 ≠ 0 𝑝𝑡(2): 𝑥2+ 𝑦2+ 2𝑦 + 1 = 2 (𝑦 + 1 − 𝑥

2

𝑥2𝑦 + 𝑦3+ 2𝑦2+ 𝑦 = 2𝑦 + 2 − 2𝑥2

𝑥2(𝑦 + 2) + 𝑦2(𝑦 + 2) − (𝑦 + 2) = 0 𝑣ậ𝑦 𝑦 = −2 ℎ𝑎𝑦 𝑥2+ 𝑦2 = 1 (∗)𝑣ớ𝑖 𝑦 = −2 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1):

4𝑥2 = (√𝑥2+ 1 + 1) 𝑥2𝑣ậ𝑦 𝑥 = 0 ℎ𝑎𝑦 𝑥 = ±2√2 (∗∗)𝑣ớ𝑖 𝑥2 = 1 − 𝑦2 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2):

4(1 − 𝑦2) = (√2 − 𝑦2+ 1) (−𝑦3 − 𝑦2+ 3𝑦 − 1)(𝑙𝑖ê𝑛 ℎợ𝑝 𝑣𝑠 𝑛𝑔ℎ𝑖ệ𝑚 𝑦 = 1)

𝑏à𝑖 35: { 𝑥

3+ 3𝑥𝑦2+ 3𝑥 = 𝑦(3𝑥2+ 𝑦2 + 3)(1)3𝑥4(𝑦2− 2) − 2𝑥3(𝑦2+ 8) + 2𝑥(15𝑦 − 6) = 0(2)

𝑔𝑖ả𝑖 𝑝𝑡(1): 𝑥3+ 3𝑥𝑦2 + 3𝑥 = 3𝑥2𝑦 + 𝑦3 + 3𝑦

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

𝑣ậ𝑦 𝑎2− 𝑏2 = 5𝑥

{ 𝑎 − 𝑏 = 4(1)2𝑏 − √𝑎2− 𝑏2+ 8 = 2(2)𝑙ấ𝑦 (1)𝑡ℎế (2): 2𝑏 − √(4 + 𝑏)2− 𝑏2+ 8 = 2

24 + 8𝑏 = 4𝑏2− 8𝑏 + 4

𝑏 = 5(𝑛ℎậ𝑛) ℎ𝑎𝑦 𝑏 = −1 (𝑙𝑜ạ𝑖) (∗)𝑣ớ𝑖 𝑏 = 5 𝑡ℎì 𝑎 = 9 (𝑡ℎõ𝑎) {2𝑥 + 𝑦 = 257𝑥 + 𝑦 = 81 => 𝑥 =56

5 , 𝑦 =

135

𝑏à𝑖 37: {𝑥 + √𝑥2 − 2𝑥 + 5

3

= 3𝑦 + √𝑦3 2+ 4(1)(𝑥 − 𝑦)(𝑥 + 𝑦 − 3) = −1(2)

𝑔𝑖ả𝑖 𝑝𝑡(1): 𝑥 + √(𝑥 − 1)3 2+ 4 = 3𝑦 + √𝑦3 2+ 4𝑝𝑡(2): 𝑥2− 𝑦2− 3𝑥 + 3𝑦 + 1 = 0 𝑙ấ𝑦 𝑝𝑡(2)𝑡ℎế 𝑝𝑡(1): 3𝑦 = 𝑦2− 𝑥2+ 3𝑥 − 1

𝑥 + √(𝑥 − 1)3 2+ 4 = 𝑦2− 𝑥2+ 3𝑥 − 1 + √𝑦3 2+ 4(𝑥 − 1)2+ √(𝑥 − 1)3 2+ 4 = 𝑦2+ √𝑦3 2+ 4

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑦 ≥ 1 𝑝𝑡(1): 𝑥3− 𝑥2+ 𝑥 = (√𝑦 − 1)3− (√𝑦 − 1)2+ (√𝑦 − 1)

𝑏à𝑖 39: {√𝑥3 3 + 2+ 𝑥2(𝑥3+ 𝑥 − 𝑦 − 1) = √𝑦 + 33 + 𝑦 + 1(1)

√4𝑥2− 2

3

− 4𝑥2− 25𝑥 − 24 = 2𝑦(2)𝑔𝑖ả𝑖

𝑝𝑡(1): √𝑥3 3+ 2− √𝑦 + 33 + 𝑥3(𝑥2+ 1) − 𝑥2(𝑦 + 1) − (𝑦 + 1) = 0

𝑥3− 𝑦 − 1(√𝑥3 3+ 2)2+ √(𝑥3 3+ 2)(𝑦 + 3)+ (√𝑦 + 33 )2

𝑙ấ𝑦 𝑝𝑡(1) 𝑡ℎế 𝑝𝑡(2): 10√(𝑥 + 7)3 2− 2𝑦3 − 2(𝑦 + 5)√(𝑥 + 7)3 2+ 4𝑦23√𝑥 + 7

= 0

−2𝑦3 − 2𝑦 √(𝑥 + 7)3 2+ 4𝑦23√𝑥 + 7 = 0

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

𝑣ậ𝑦 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓(√𝑥 − 13 ) = 𝑓(𝑦2+ 1) 𝑣ậ𝑦 √𝑥 − 13 = 𝑦2+ 1 (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) đế𝑛 đâ𝑦 𝑑à𝑛ℎ 𝑐ℎ𝑜 𝑏ạ𝑛 đọ𝑐)

𝑏à𝑖 45: {[𝑥

2+ 2𝑦 − 3 − 3√𝑦(𝑥2+ 1) − (𝑦2+ 𝑥2)] √𝑥2− 𝑦 = √(𝑦 − 1)3(1)

𝑦 − 𝑥 = √2𝑥2+ 1(2)𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ { 𝑦 ≥ 1

𝑥2− 𝑦 ≥ 0𝑝𝑡(1): [𝑥2+ 2𝑦 − 3 − 3√𝑦𝑥2+ 𝑦 − 𝑦2− 𝑥2] √𝑥2− 𝑦 = √(𝑦 − 1)3

[𝑥2+ 2𝑦 − 3 − 3√(𝑥2− 𝑦)(𝑦 − 1)] √𝑥2− 𝑦 = √(𝑦 − 1)3

đặ𝑡 𝑎(𝑥2− 𝑦) + 𝑏(𝑦 − 1) = 𝑥2+ 2𝑦 − 3 {

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: {

22𝑥+3 + 9 ≥ 03.2𝑦+1 − 1 ≥ 0

22𝑥+1 + 3 ≥ 03.2𝑦+3 − 7 ≥ 0𝑝𝑡(1): √8.22𝑥 + 8 + 1 + √(3.2𝑦) 2 − 2 + 1

= √22𝑥 2 + 2 + 1 + √(3.2𝑦) 8 − 8 + 1

√8(22𝑥 + 1) + 1 + √2(3.2𝑦 − 1) + 1 = √2(22𝑥 + 1) + 1 + √8(3.2𝑦 − 1) + 1

𝑥é𝑡 𝑓(𝑡) = √8𝑡 + 1 − √2𝑡 + 1 𝑣ậ𝑦 𝑓′(𝑡) > 0 𝑛ê𝑛 𝑡𝑎 𝑐ó 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑣ậ𝑦 22𝑥 + 1 = 3.2𝑦 − 1(∗∗) 𝑝𝑡(2): √𝑥3 2+ 𝑥 + 𝑦2− 𝑦 − 2𝑥𝑦 + 1

√𝑡2+ 𝑡 + 1

3

= 𝑡3+ 2𝑡2 + 3𝑡 + 1 𝑣ậ𝑦 𝑡𝑎 𝑡ì𝑚 đượ𝑐 𝑡 = 0 𝑣ậ𝑦 𝑥 = 𝑦 𝑡ℎế 𝑥 = 𝑦 𝑣à𝑜 𝑝𝑡(∗∗)22𝑥 − 3.2𝑥+ 2 = 0

𝑥2+ 4𝑦 + 1 ≥ 0

𝑝𝑡(1): 𝑥3− 3𝑥2+ 6𝑥 + √𝑥 − 1 = 8𝑦3− 12𝑦2+ 12𝑦 + √2𝑦 − 1

𝑥3− 3𝑥2+ 3𝑥 − 1 + 3𝑥 + √𝑥 − 1 = 8𝑦3− 12𝑦2+ 6𝑦 − 1 + 6𝑦 + √2𝑦 − 1 (𝑥 − 1)3+ 3(𝑥 − 1) + √𝑥 − 1 = (2𝑦 − 1)3+ 3(2𝑦 − 1) + √2𝑦 − 1

𝑥é𝑡 𝑓(𝑡) = 𝑡3+ 3𝑡 + √𝑡 𝑣ậ𝑦 𝑓′(𝑡) = 3𝑡2+ 3 + 1

2√𝑡 > 0 𝑣ậ𝑦 𝑥 − 1 = 2𝑦 − 1 𝑣ậ𝑦 𝑥 = 2𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)

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

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

𝑣ậ𝑦 𝑥 = −1 ℎ𝑎𝑦 𝑥 = 𝑦 + 3

𝑏à𝑖 51: { (𝑦 − 1) (√𝑥2− 1 − √𝑦 − 1) = 2(1)

(𝑥2+ 4𝑦)√𝑥2− 1 + 6 = 5√𝑥2− 1 (1 + √(𝑥2− 1)(𝑦 − 1)) (2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: {|𝑥| ≥ 1

𝑦 ≥ 1𝑝𝑡(2): (𝑥2− 1)√𝑥2− 1 + 4(𝑦 − 1)√𝑥2− 1 + 6 = 5 (√𝑥2− 1)2√𝑦 − 1

𝑡ℎế 𝑝𝑡(1)𝑣à𝑜 𝑝𝑡(2):

(√𝑥2− 1)3+ 4(√𝑦 − 1)2√𝑥2− 1 − 5 (√𝑥2 − 1)2√𝑦 − 1

+ 3(√𝑦 − 1)2√𝑥2− 1 − 3(√𝑦 − 1)3 = 0 (√𝑥2− 1)3+ 7(√𝑦 − 1)2√𝑥2− 1 − 5 (√𝑥2− 1)2√𝑦 − 1 − 3(√𝑦 − 1)3 = 0

𝑣ậ𝑦 √𝑥2− 1 = 3√𝑦 − 1 ℎ𝑎𝑦 √𝑥2− 1 = √𝑦 − 1 (∗)𝑣ớ𝑖 √𝑥2− 1 = 3√𝑦 − 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1):

2(𝑦 − 1) = 2 𝑣ậ𝑦 𝑦 = 2 𝑡ℎì 𝑥 = ±√10 (∗∗)𝑣ớ𝑖 √𝑥2− 1 = √𝑦 − 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1):

(𝑝𝑡𝑣𝑛)

𝑏à𝑖 52: {

2𝑥3+ 2𝑥 = 𝑦3− 𝑦2+ 3𝑥𝑦(𝑥 − 𝑦) + 𝑦(2𝑥 + 1)(1)(𝑦 + 1) (2 + 2√𝑥2+ 1 +𝑦

(𝑦 − 𝑥)3− 𝑥3− 𝑦(𝑦 − 2𝑥) + 𝑦 − 2𝑥 = 0 (𝑦 − 𝑥)3− 𝑦2+ 2𝑥𝑦 − 𝑥2+ 𝑦 = 𝑥3+ 2𝑥 − 𝑥2(𝑦 − 𝑥)3− (𝑦 − 𝑥)2+ (𝑦 − 𝑥) = 𝑥3− 𝑥2 + 𝑥 𝑥é𝑡 𝑓(𝑡) = 𝑡3− 𝑡2 + 𝑡 𝑣ậ𝑦 𝑓′(𝑡) = 3𝑡2− 2𝑡 + 1 > 0 𝑛ê𝑛 𝑦 = 2𝑥

> 0 𝑛ê𝑛 𝑥2 = 𝑦 − 𝑥 (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)𝑏ì𝑛ℎ 𝑡ℎườ𝑛𝑔)

𝑏à𝑖 54: {3𝑥

3+ 7𝑥 − 9𝑦 = (3𝑦 + 13)√𝑦 − 1 + 1(1)2𝑦 + 6𝑥 + 11 = (4𝑥 − 1)√𝑥2+ 8(2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑦 ≥ 1

𝑝𝑡(1): 3(𝑥 − 1)3+ 9𝑥2− 18𝑥 + 9 + 16(𝑥 − 1)

= 3(√𝑦 − 1)3+ 9(√𝑦 − 1)2+ 16(√𝑦 − 1) 3(𝑥 − 1)3+ 9(𝑥 − 1)2 + 16(𝑥 − 1)

= 3(√𝑦 − 1)3+ 9(√𝑦 − 1)2+ 16(√𝑦 − 1) 𝑥é𝑡 𝑓(𝑡) = 3𝑡3+ 9𝑡2+ 16𝑡 𝑣ậ𝑦 𝑓′(𝑡) = 9𝑡2+ 18𝑡 + 16 > 0

𝑣ậ𝑦 𝑥 − 1 = √𝑦 − 1(𝑡ℎế 𝑣à𝑜 𝑝𝑡(2))

𝑏à𝑖 55: {

3𝑥3+ 𝑥2 + 𝑥 = 10𝑦 + 12√𝑦 + 3(√𝑦)3+ 5(1)4(2𝑥2+ 1) + 3(𝑥2− 2𝑥)√1 + 2√𝑦 = 2(𝑥3 + 5√𝑦 + 5)(2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑦 ≥ 0 𝑝𝑡(1): 3𝑥3+ 𝑥2+ 𝑥 = 3(√𝑦)3+ 9(√𝑦)2+ 9√𝑦 + 3 + 𝑦 + 3√𝑦 + 2

2+ 3𝑦 = 9(1)

𝑦4+ 4(2𝑥 + 3)𝑦2− 48𝑦 + 48𝑥 + 155 = 0(2)

𝑔𝑖ả𝑖 𝑙ấ𝑦 𝑝𝑡(1)𝑡ℎế 𝑣à𝑜 𝑝𝑡(2):

𝑦4+ 4𝑦2(2𝑥 + 3) − 16(9 − 𝑥2) + 48𝑥 + 155 = 0

= 12(2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 𝑥 ≥ 𝑦 𝑝𝑡(2): √(𝑥 − 𝑦)(𝑥8 2 + 𝑥𝑦 + 𝑦2) + 𝑥𝑦(𝑥 − 𝑦) = 12

√(𝑥 − 𝑦)(𝑥 + 𝑦)2 8

{𝑎 + 𝑏 = 8

𝑎 𝑏 = 12

𝑏à𝑖 61: {𝑥 + √𝑥(𝑥

2− 6𝑥 + 12) = √𝑦 + 23 + √𝑦 + 10 + 2(1)3√𝑥 − 1 − √𝑥2− 6𝑥 + 6 = √𝑦 + 23 + 2(2)

3√𝑥 − 1 − √𝑥2− 6𝑥 + 6 = 𝑥(𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = 5 , 𝑥 = 1.25 𝑠𝑜 𝑙ạ𝑖 𝑣ớ𝑖 đ𝑘)

𝑏à𝑖 62: { (4𝑦 − 1)√𝑥2+ 4 = 2𝑥2+ 2𝑦 + 7(1)

𝑥5 − 3(4𝑦2− 4𝑦 − 3)2− 𝑥3+ 4(3𝑥2− 𝑦2) − 4(3𝑥 − 𝑦) + 7 = 0(2)

𝑔𝑖ả𝑖

𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ℎ𝑜 2 𝑐ủ𝑎 𝑝𝑡(1): (2𝑦 −1

2) √𝑥

2+ 4 = 𝑥2+ 𝑦 +7

2𝑝𝑡(1): (√𝑥2+ 4 + 𝑎) (√𝑥2+ 4 + 𝑏) = 0

= 0 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1) 𝑡ℎấ𝑦 𝑘ℎô𝑛𝑔 𝑡ℎõ𝑎 𝑇𝐻2: 𝑛ế𝑢 𝑦 ≠ 0 𝑡ℎì 𝑐ℎ𝑖𝑎 2 𝑣ế 𝑐ủ𝑎 𝑝𝑡(2)𝑐ℎ𝑜 𝑦:

𝑏à𝑖 65: {𝑥

2+ 𝑥 = 𝑦3− 𝑦(1)

𝑦2 + 𝑦 = 𝑥3− 𝑥(2)𝑔𝑖ả𝑖

{𝑥(𝑥 + 1) = 𝑦(𝑦 − 1)(𝑦 + 1)(1)𝑦(𝑦 + 1) = 𝑥(𝑥 − 1)(𝑥 + 1)(2)𝑙ấ𝑦 𝑝𝑡(1) 𝑝𝑡(2): 𝑥𝑦(𝑥 + 1)(𝑦 + 1) = 𝑥𝑦(𝑥 + 1)(𝑦 + 1)(𝑥 − 1)(𝑦 − 1)

𝑣à 𝑥𝑦 − 𝑥 − 𝑦 = 0 𝑙ấ𝑦 𝑝𝑡(1) + 𝑝𝑡(2): 𝑥2 + 𝑦2+ 2(𝑥 + 𝑦) = (𝑥 + 𝑦)(𝑥2− 𝑥𝑦 + 𝑦2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≥ 1 (𝑑𝑜 𝑝𝑡(2)) 𝑣à 𝑦 ≥ 1

𝑝𝑡(1): 3√4𝑥(𝑥 + 1)3 + 2√2𝑦2− 𝑦 ≤ 2 + 2𝑥 + 𝑥 + 1 + 𝑦 + 2𝑦 − 1

= 3(𝑥 + 𝑦) + 2 𝑑ấ𝑢 "=" xảy ra khi : {𝑥 = 1𝑦 = 1(𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎)

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ {𝑥𝑦 ≥ 0

𝑥 ≥ −1

để ý 𝑝𝑡(1): 16𝑥 + 16 = 4(𝑥𝑦)2+ 𝑥2𝑦4(4 − 𝑥𝑦2)(4 + 𝑥𝑦2) + 4𝑥(4 − 𝑥𝑦2) = 0 𝑣ậ𝑦 𝑥𝑦2 = 4 ℎ𝑎𝑦 4 + 𝑥𝑦2+ 4𝑥 = 0

𝑥𝑦2 = 4 ℎ𝑎𝑦 𝑥(4 + 𝑦2) = −4 => 𝑦2 = −4

𝑥 − 4(𝑣𝑠 𝑥 ≠ 0)

𝑥𝑦 + (𝑥𝑦 − 2)2𝑥𝑦 + 𝑥𝑦 = 3(1)(𝑥 + 𝑥𝑦 + 2)(3 𝑥4+ 8𝑥2+ 12𝑥 + 𝑥𝑦 + 20) = 5(𝑥2+ 𝑥𝑦)3(2)

𝑔𝑖ả𝑖 𝑝𝑡(1): (2𝑥𝑦)2+ (𝑥𝑦 − 2) 2𝑥𝑦 + (𝑥𝑦 − 2) − 1 = 0

(2𝑥𝑦 − 1)(2𝑥𝑦+ 1) + (𝑥𝑦 − 2)(2𝑥𝑦 + 1) = 0 𝑣ậ𝑦 2𝑥𝑦 + 𝑥𝑦 − 3 = 0 ℎ𝑎𝑦 2𝑥𝑦 = −1(𝑙𝑜ạ𝑖)

𝑥𝑦 = 1 (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2))

𝑏à𝑖 69: {

𝑥√4𝑦2+ 1 + 2𝑦√𝑥2+ 1 = 0(1)2(1 + 23𝑥−4𝑦)

1 + 2−2𝑦 + 2.2

3𝑥

1 + 2−4𝑦 = 2−2𝑦(1 + 2−2𝑦 + 22𝑥)(2)𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≥ 0, 𝑦 ≥ 0 4(𝑥𝑦)2+ 𝑥2 = 4(𝑥𝑦)2+ 4𝑦2

𝑣ậ𝑦 𝑥 = ±2𝑦 (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2))

𝑏à𝑖 70: {2𝑥(4𝑥

2− 𝑦 + 2) + 4𝑥2(𝑦 − 1) = √𝑦 − 2 − 2𝑥 + 𝑦2− 3𝑦 + 2(1)(√𝑦 − 2 − 1)√2𝑥 + 1 = 8𝑥3− 13(𝑦 − 2) + 82𝑥 − 29(2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: 2𝑥 + 1 ≥ 0, 𝑦 − 2 ≥ 0 𝑝𝑡(1): 2𝑥(4𝑥2− 𝑦 + 2) + 4𝑥2(𝑦 − 1) = √𝑦 − 2 − 2𝑥 + (𝑦 − 2)(𝑦 − 1)

2𝑥(4𝑥2− 𝑦 + 2) + (𝑦 − 1)(4𝑥2− 𝑦 + 2) = √𝑦 − 2 − 2𝑥

(2𝑥 + 𝑦 − 1)(4𝑥2− 𝑦 + 2) = −(4𝑥

2− 𝑦 + 2)

√𝑦 − 2 + 2𝑥𝑣ậ𝑦 4𝑥2− 𝑦 + 2 = 0 ℎ𝑎𝑦 2𝑥 − 𝑦 + 1 = (2𝑥 + 1) + (𝑦 − 2)

√𝑦 − 2 + 2𝑥(𝑝𝑡𝑣𝑛 𝑑𝑜 đ𝑘) 𝑣ớ𝑖 4𝑥2− 𝑦 + 2 = 0 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)𝑔𝑖ả𝑖 𝑛ó 𝑡ℎ𝑒𝑜 𝑝𝑝 đ𝑜á𝑛 𝑛𝑔ℎ𝑖ệ𝑚

𝑏à𝑖 71: {20√6 − 𝑥 − 17√5 − 𝑦 − 3𝑥√6 − 𝑥 + 3𝑦√5 − 𝑦 = 0(1)

2√2𝑥 + 𝑦 + 5 + 3√3𝑥 + 2𝑦 + 11 = 𝑥2+ 6𝑥 + 13(2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≤ 6, 𝑦 ≤ 5 𝑣à {3𝑥 + 2𝑦 + 11 ≥ 02𝑥 + 𝑦 + 5 ≥ 0𝑝𝑡(1): (20 − 3𝑥)√6 − 𝑥 = (17 − 3𝑦)√5 − 𝑦 2√6 − 𝑥 + 3(√6 − 𝑥)3 = 2√5 − 𝑦 + 3(√5 − 𝑦)3𝑥é𝑡 𝑓(𝑡) = 2𝑡 + 3𝑡3 𝑣ậ𝑦 𝑓′(𝑡) = 2 + 9𝑡2 > 0 𝑛ê𝑛 6 − 𝑥 = 5 − 𝑦 𝑣ậ𝑦 𝑥 − 𝑦 = 1 (𝑡ℎế 𝑣à𝑜 (2))

𝑏à𝑖 73: {

12𝑥2− 𝑦(2𝑥2− 4𝑥 + 3) − 12𝑥 + 3 = 0(1)512

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≥ −1 𝑇𝐻1: 𝑛ế𝑢 𝑥 = 0 𝑡ℎì 𝑝𝑡𝑣𝑛

𝑥 − 2(2𝑦 + 1)

√𝑥 − (2𝑦 + 1) + √2𝑦 + 1 =

−3(𝑥 − 2(2𝑦 + 1))

√𝑥 + 14(2𝑦 + 1) + √4𝑥 + 8(2𝑦 + 1)𝑣ậ𝑦 𝑥 − 2(2𝑦 + 1) = 0 ℎ𝑎𝑦 𝑝𝑡 𝑐ò𝑛 𝑙ạ𝑖 𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚 𝑑𝑜 𝑉𝑇 > 0 𝑚à 𝑉𝑃 < 0

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑦 ≥ −4 𝑣à 𝑥 ≤ 2 𝑝𝑡(1): (4𝑥 + 2)√𝑥2+ 𝑥 + 1 + (2𝑥 − 𝑦)√(2𝑥 − 𝑦)2+ 3 = 𝑦 − 4𝑥 − 1 (2𝑥 + 1)√(2𝑥 + 1)2+ 3 + (2𝑥 + 1) = (𝑦 − 2𝑥)√(𝑦 − 2𝑥)2+ 3 + (𝑦 − 2𝑥)

𝑥é𝑡 𝑓(𝑡) = 𝑡√𝑡2+ 3 + 𝑡 𝑣ậ𝑦 𝑓′(𝑡) > 0 𝑛ê𝑛 𝑡𝑎 𝑐ó 2𝑥 + 1 = 𝑦 − 2𝑥 => 4𝑥 + 1

= 𝑦

𝑏à𝑖 77: {2𝑥

2𝑦 + 3𝑥𝑦 = 4𝑥2+ 9𝑦(1)7𝑦 + 6 = 2𝑥2+ 9𝑥(2)𝑔𝑖ả𝑖

𝑝𝑡(∗∗): 𝑥 = −2, 𝑥 =1

2, 𝑥 =

−9 ± 3√33

4 (𝑡ℎế 𝑡ừ𝑛𝑔 𝑐á𝑖 𝑣à𝑜 𝑡𝑎 𝑡ì𝑚 đượ𝑐 𝑦) 𝑏à𝑖 78: {𝑥

2+ 𝑦2+ 𝑧2 = 20102(1)

𝑥3+ 𝑦3+ 𝑧3 = 20103(2)

𝑔𝑖ả𝑖

𝑡ừ 𝑝𝑡(1): 𝑡𝑎 𝑠𝑢𝑦 𝑟𝑎 đượ𝑐 |𝑥|, |𝑦|, |𝑧| ≤ 2010 𝑠𝑢𝑦 𝑟𝑎 𝑥3+ 𝑦3+ 𝑧3 ≤ |𝑥3| + |𝑦3| + |𝑧3| ≤ 2010(𝑥2+ 𝑦2+ 𝑧2) = 20103

2+ 1

𝑦 , 𝑏 = 𝑥 + 𝑦 {

+ √6𝑥 − 1 = √2𝑦 − 1 + 1

− (𝑦𝑥)3+ 3 (𝑦𝑥)2(√1 − (3 𝑦𝑥)3 + 3 (𝑦𝑥)2)

2

+ √1 − (3 𝑦𝑥)3+ 3 (𝑦𝑥)2 + 1

√2𝑦 − 1 + √6𝑥 − 1

− (𝑦𝑥)2(𝑦𝑥 − 3)(√1 − (3 𝑦𝑥)3 + 3 (𝑦𝑥)2)

𝑏à𝑖 81: { √𝑥4 + 𝑥2+ 𝑥 − 𝑦 = 𝑦 − 𝑥(1)

√𝑥2+ 𝑥 − 2 + (𝑦 − 1)√𝑦 + 6 + (𝑥2+ 𝑥 − 3)(𝑦 − 1) = 7(2)

𝑔𝑖ả𝑖 𝑝𝑡(1): 𝑥4+ 𝑥2+ 𝑥 − 𝑦 = 𝑦2− 2𝑥𝑦 + 𝑥2(𝑦 ≥ 𝑥)

𝑥4+ 𝑥 − 𝑦 = 𝑦2 − 2𝑥𝑦

𝑦2+ 𝑦(−2𝑥 + 1) − 𝑥4 − 𝑥 = 0 𝑑𝑒𝑛𝑡𝑎 = 4𝑥2− 4𝑥 + 1 − 4(−𝑥4− 𝑥) = 4𝑥2+ 1 + 4𝑥4 = (2𝑥2+ 1)2 ≥ 0

+ 𝑥 − 𝑦 = 4(1)𝑥√𝑥 − 𝑦 − (𝑥2+ 1)√𝑦 + (𝑥3 + 2)√𝑦 + 3 = 17(2)

𝑥 = 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 > 0 ) 𝑣ậ𝑦 √(𝑥 + 2𝑦)𝑥 = √𝑥3 2(𝑥 + 3𝑦)

(𝑥2+ 2𝑦𝑥)3 = 𝑥4(𝑥 + 3𝑦)2

𝑥6+ 6𝑥5𝑦 + 12𝑥4𝑦2+ 8(𝑥𝑦)3 = 𝑥4(𝑥2+ 6𝑥𝑦 + 9𝑦2) = 𝑥6+ 6𝑥5𝑦 + 9𝑥4𝑦2

𝑏à𝑖 85: { 4𝑥𝑦 + 1 = 𝑥 + 2√𝑥𝑦(1)

(𝑥√𝑥)−1 + 8𝑦√𝑦 = (√𝑥)−1 + 6√𝑦(2)

𝑔𝑖ả𝑖

đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 > 0 𝑣à 𝑦 ≥ 0

(𝑦√2 − 𝑥)3− 8 + 3𝑦3√2 − 𝑥 − 6𝑦2 = 0 (𝑦√2 − 𝑥 − 2)(𝑦2(2 − 𝑥) + 2𝑦√2 − 𝑥 + 4) + 3𝑦2(𝑦√2 − 𝑥 − 2) = 0

𝑣ậ𝑦 𝑦√2 − 𝑥 − 2 = 0 ℎ𝑎𝑦 𝑦2(2 − 𝑥) + 2𝑦√2 − 𝑥 + 4 + 3𝑦2 = 0 (𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇

> 0) 𝑣ớ𝑖 𝑦√2 − 𝑥 − 2 = 0 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) 𝑡𝑎 đượ𝑐 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = 1.2

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

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

+ 2(√𝑥 − 2 − √𝑦 + 1) − √4𝑦 + 4 + √𝑥 + 3𝑦 + 1 = 0 (𝑥 − 𝑦 − 3) ((√𝑥 − 2)2+ √(𝑥 − 2)(𝑦 + 1) + (√𝑦 + 1)2)

2(√𝑥 − 2 + √𝑦 + 1)+

1(√𝑥 + 3𝑦 + 1 + √4𝑦 + 4)= 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 > 0)