tuyển tập lời giải hay

Chào Thọ Nguyễn Đức. Nhằm mục tiêu phát triển bền vững. Từ tháng 082016, BQT 123doc.org kêu gọi thành viên chưa chia sẻ tài liệu trong năm vừa qua chia sẻ ít nhất 2 tài liệu miễn phínăm cho cộng đồng. Đây là lời kêu gọi dành cho các bạn biết cảm ơn những giá trị đã nhận được từ 123doc.org, và có trách nhiệm duy trì giá trị đó với cộng đồng. Bạn hãy UPLOAD chia sẻ ngay để trở thành một phần của lý tưởng tốt đẹp này

+ √1 − 𝑦 − 2

𝑓′(𝑦) = −3𝑦2

3 √(10−𝑦3 3 ) 2+ −1

2√1−𝑦> 0(đ𝑘 𝑦 < 1) ℎà𝑚 𝑛à𝑦 𝑙𝑢ô𝑛 đồ𝑛𝑔 𝑏𝑖ế𝑛 𝑣ậ𝑦 𝑝ℎả𝑖 𝑐ó í𝑡 𝑛ℎấ𝑡 1 𝑛𝑔ℎ𝑖ệ𝑚 𝑓(𝑦) = 0

𝑦+ 𝑦 = 3 ℎ𝑎𝑦 𝑥

𝑦+ 𝑦 = −1 𝑣ớ𝑖𝑥

=> 6𝑦2 − 18𝑦 + 8 = 0 => 𝑦 =9+√33

6 (𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑦 = 9−√33

6 (𝑙𝑜ạ𝑖) 𝑣ớ𝑖𝑥

=> 𝑦4 + 2𝑦3+ 𝑦2 + 1 = 𝑦4 + 2𝑦3− 𝑦2 − 2𝑦 + 1

=> 2𝑦2 + 2𝑦 = 0 => 𝑦 = −1(𝑡ℎõ𝑎)

𝑦 = −1 => 𝑥 = 0

3) {𝑥(𝑥𝑦 + 𝑥)

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

𝑝𝑡(1): 𝑥(𝑥𝑦 + 𝑥)2+ (𝑥 + 1)2 = 𝑥3(𝑦2+ 𝑦 + 1) + 𝑥2(𝑦 − 1) + 5𝑥

=> 𝑥(𝑥2𝑦2+ 2𝑥2𝑦 + 𝑥2) + (𝑥2 + 2𝑥 + 1) = 𝑥3(𝑦2+ 𝑦 + 1) + 𝑥2(𝑦 − 1) + 5𝑥 => 𝑥3𝑦2+ 2𝑥3𝑦 + 𝑥3+ 𝑥2+ 2𝑥 + 1 = 𝑥3𝑦2+ 𝑥3𝑦 + 𝑥3+ 𝑥2𝑦 − 𝑥2 + 5𝑥 => 𝑥3𝑦 + 2𝑥2 − 3𝑥 + 1 = 𝑥2𝑦

𝑥 3+21𝑥 𝑥é𝑡 𝑓(𝑡) = 𝑡3+ 21𝑡 => 𝑓′(𝑡) = 3𝑡2+ 21 > 0

𝑣ậ𝑦 𝑦 =1

𝑥 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): 21𝑦 − 20 − 1

𝑥 3 = 0 => 21𝑦 − 20 − 𝑦3 = 0 => 𝑦 = −5 ℎ𝑎𝑦 𝑦 = 4 ℎ𝑎𝑦 𝑦 = 1

=> 𝑥2+ 𝑥𝑦 − 2𝑦2+ 𝑥−𝑦

√𝑥+𝑦+√2𝑦= 0 => (𝑥 − 𝑦)(𝑥 + 2𝑦) + 𝑥−𝑦

√𝑥+𝑦+√2𝑦= 0

𝑣ậ𝑦 𝑥 = 𝑦 ℎ𝑎𝑦 𝑥 + 2𝑦 + 1

√𝑥+𝑦+√2𝑦= 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 > 0) 𝑣ớ𝑖 𝑥 = 𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): 8𝑥2− 8𝑦 + 2(𝑥3− 𝑦3) + 3 = 8𝑦√2𝑥2 − 3𝑥 + 1 => 8𝑥2− 8𝑥 + 3 = 8𝑥√2𝑥2− 3𝑥 + 1(đ𝑘: 2𝑥2− 3𝑥 + 1 ≥ 0)

=> 8𝑥2− 11𝑥 + 3 = 8𝑥√2𝑥2− 3𝑥 + 1 −8𝑥

2 => 8𝑥2− 11𝑥 + 3 = 8𝑥(2𝑥

2 −3𝑥+3

4 )

√2𝑥 2 −3𝑥+1+12 𝑣ậ𝑦 8𝑥2− 11𝑥 + 3 = 0 ℎ𝑎𝑦 1 = 2𝑥

√2𝑥 2 −3𝑥+1+12

8𝑥2− 11𝑥 + 3 = 0 => 𝑥 = 𝑦 = 1 ℎ𝑎𝑦 𝑥 = 𝑦 = 3

8 √2𝑥2− 3𝑥 + 1 +1

2 = 2𝑥 => √2𝑥2 − 3𝑥 + 1 = 2𝑥 −1

2(đ𝑘 𝑥 ≥1

4) => 2𝑥2− 3𝑥 + 1 = 4𝑥2− 2𝑥 +1

4

=> 𝑥 = 𝑦 =−1+√7

4 ℎ𝑎𝑦 𝑥 =−1−√7

4 (𝑙𝑜ạ𝑖) 6) {𝑥

𝑦 = − 1

√3 3 𝑣ớ𝑖 𝑏 = 1 => 𝑎 = √33

𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖: {𝑥 + 𝑦 = √33

𝑥 − 𝑦 = 1 => {

𝑥 = √3

3 +1 2

𝑦 = √3

3

−1 2

7) {𝑥

2 + 𝑦2+ 𝑥𝑦 = 4𝑦 − 1(1)

𝑥 + 𝑦 − 2 = 𝑦

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

=> 𝑥4−4𝑥2(𝑥+𝑥2)

2𝑥−1 + 3𝑥2+ (𝑥+𝑥2

2𝑥−1)2 = 0 => 𝑥4(2𝑥 − 1)2− 4𝑥2(𝑥2+ 𝑥)(2𝑥 − 1) + 3𝑥2(2𝑥 − 1)2+ (𝑥 + 𝑥2)2 = 0 => 2𝑥2(𝑥 − 2)(𝑥 − 1)(2𝑥2+ 1) = 0

=> 𝜕𝑡2− 𝑡(6𝑦 − 4𝑥 + 6) + (2𝑥 − 5𝑦 + 4) − 𝜕(𝑥 − 𝑦 − 1) = 0

𝑃 = √𝑑𝑒𝑛𝑡𝑎 = √(6𝑦 − 4𝑥 + 6)2− 4𝜕[(2𝑥 − 5𝑦 + 4) − 𝜕(𝑥 − 𝑦 − 1)] 𝑐ℎ𝑜 𝑥 = 100, 𝑦 = 1000 𝑣à 𝑐ℎọ𝑛 𝑠ố 𝜕 𝑠𝑎𝑜 𝑐ℎ𝑜 𝑃 𝑙à 𝑠ố 𝑛𝑔𝑢𝑦ê𝑛

=> 𝑥2+ 𝑥 =3√−𝑥4−2𝑥3+5𝑥2+6𝑥+4−6

√−𝑥 4 −2𝑥 3 +5𝑥 2 +6𝑥+4

=> 𝑥(𝑥 + 1)√−𝑥4− 2𝑥3+ 5𝑥2+ 6𝑥 + 4 = 3(−𝑥4−2𝑥3+5𝑥2+6𝑥)

√−𝑥 4 −2𝑥 3 +5𝑥 2 +6𝑥+4+2 => 𝑥(𝑥 + 1)√−𝑥4− 2𝑥3+ 5𝑥2+ 6𝑥 + 4 = −3𝑥(𝑥+1)(𝑥−2)(𝑥+3)

√−𝑥 4 −2𝑥 3 +5𝑥 2 +6𝑥+4+2 𝑣ậ𝑦 𝑥 = 0 ℎ𝑎𝑦 𝑥 = −1 ℎ𝑎𝑦 √– 𝑥4− 2𝑥3+ 5𝑥2+ 6𝑥 + 4 = √−𝑥4−3(𝑥−2)(𝑥+3)

−2𝑥 3 +5𝑥 2 +6𝑥+4+2 => −𝑥4 − 2𝑥3 + 5𝑥2+ 6𝑥 + 4 + 3(𝑥 − 2)(𝑥 + 3) + 2√−𝑥4− 2𝑥3 + 5𝑥2+ 6𝑥 + 4 = 0 => −(𝑥2+ 𝑥)2+ 9(𝑥2+ 𝑥) − 8 + 2√−𝑥4 − 2𝑥3+ 5𝑥2+ 6𝑥 + 4 − 6 = 0

13) {7𝑦 = 4𝑥

3+ 3𝑥𝑦2(1)7𝑥 = 𝑦3+ 6𝑥2𝑦(2)

𝑙ấ𝑦 𝑝𝑡(1) + 𝑝𝑡(2): 7(𝑥 + 𝑦) = 4𝑥3+ 6𝑥2𝑦 + 3𝑥𝑦2+ 𝑦3 = (𝑥 + 𝑦)(4𝑥2+ 2𝑥𝑦 + 𝑦2) 𝑣ậ𝑦 𝑥 + 𝑦 = 0 ℎ𝑎𝑦 4𝑥2+ 2𝑥𝑦 + 𝑦2 = 7

𝑝𝑡(1): 𝑥2+ 𝑦2+ 1

𝑥 2+ 1

𝑦 2 = 5

𝑦 +2𝑦

𝑥 − 2

𝑥𝑦= 4 𝑐ộ𝑛𝑔 2 𝑣ế 𝑐ủ𝑎 ℎệ: 𝑥2+ 1

16) {(𝑥 − 2)4 + 4(𝑥2+ 𝑥 + 𝑦)4 = 5(𝑥3− 4𝑥 − 𝑦 + 1)2(1)

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

17) {

√𝑥 1+√1−𝑥+ 𝑦 =√1−𝑦

1+√𝑦+ 1 − 𝑥(1)(𝑥+3

√𝑥𝑦+√(1−𝑦)(1−𝑥)+ (1 + √1 − 𝑥)(1 + √𝑦) = 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 >0)

=> 𝑦 = −2𝑥 ℎ𝑎𝑦 𝑦 = −𝑥 ℎ𝑎𝑦 𝑦 = −𝑥

2𝑣ớ𝑖 𝑦 = −2𝑥 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): 𝑥2+ 𝑥𝑦 + 𝑦2 = 3

𝑥 2= 2𝑦 +4

𝑥(2) 𝑝𝑡(2): (𝑥𝑦 + 2)2+ 1

𝑥 2= 2𝑦 +4

𝑥 => 𝑥2𝑦2+ 4𝑥𝑦 + 4 + 1

𝑥=> (𝑎2− 1)3 + 𝑎 −1

2= (𝑎 − 2)3𝑎3 => 6𝑎5− 15𝑎4+ 8𝑎3+ 3𝑎2+ 𝑎 −3

2 = 0 đặ𝑡 𝑓(𝑎) = 6𝑎5− 15𝑎4 + 8𝑎3+ 3𝑎2+ 𝑎 −3

2 => 𝑓′(𝑎) = 30𝑎4− 60𝑎3+ 24𝑎2+ 6𝑎 + 1

𝑡𝑎 𝑐ó 𝑓′(𝑎) = 30(𝑎 − 𝑎2)2+ 6(𝑎 − 𝑎2) + 1 > 0

𝑣ậ𝑦 𝑝𝑡 𝑛à𝑦 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑑𝑢𝑦 𝑛ℎấ𝑡 𝑡𝑟ê𝑛 𝑅 => 𝑓 (1

2) = 0 => 𝑎 =1

2 => 𝑥 = 2 => 𝑦 = −3

4 20) { 𝑥

=> (𝑥8 + 𝑦8− 2𝑥4𝑦4) + 2𝑥4𝑦4− 4𝑥2𝑦2+ 2 = 0

𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): ( −𝑥

𝑥 2 −6)2(𝑥2− 3) − 𝑥 ( −𝑥

𝑥 2 −6) + 1 = 0 => 𝑥 = ±√3 ℎ𝑎𝑦 𝑥 = ±2 𝑣ớ𝑖 𝑥 = ±2 => 𝑦 = ∓1

=> (𝑦 + 1)2 = 𝑦4 => 𝑦 =1±√5

2 𝑣ớ𝑖 𝑥𝑦 =1

3 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1):

=> (𝑦

3+ 1)2 = −𝑦4

3 (𝑝𝑡𝑣𝑛) 24) { 𝑥

26) {2𝑥

3 + 𝑥(𝑦2+ 1) = 3𝑦3+ 13𝑦2 + 𝑥2 + 21𝑦 + 14(1)

√17 − (𝑦 + 2)8 4

2 , 𝑏 = √𝑥−𝑦2 (đ𝑘: 𝑎, 𝑏 ≥ 0) => 2𝑎2 = 𝑥 + 𝑦, 2𝑏2 = 𝑥 − 𝑦

5 1− √25

−1

√92

5 +1

𝑦= −6 ℎ𝑎𝑦 𝑥2+1

𝑦 = 3 𝑣ớ𝑖 𝑥2+1

(3𝑥−2𝑦)(𝑥+3𝑦)= 2 => 𝑥 + 3𝑦 + 1

𝑥+3𝑦= 2 => (𝑥 + 3𝑦)2− 2(𝑥 + 3𝑦) + 1 = 0 => 𝑥 + 3𝑦 = 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1)

(𝑥−2𝑦)(3𝑥−𝑦)= 1 => 𝑥 − 2𝑦 − 2

𝑥−2𝑦= 1 => (𝑥 − 2𝑦)2− (𝑥 − 2𝑦) − 2 = 0 𝑣ậ𝑦 𝑥 − 2𝑦 = 2 ℎ𝑎𝑦 𝑥 − 2𝑦 = −1

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

=> 𝑦 + 3𝑦+2

10𝑦+12= 2 => 𝑦 = 1±√

181 5

4 => 𝑥 =25±√905

10 𝑣ớ𝑖 𝑥 − 2𝑦 = −1 => 𝑥 = 2𝑦 − 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1):

=> 2𝑧3+ 2𝑥2𝑧 + 2𝑦2𝑧 − 4𝑥𝑦𝑧 − (𝑥3 + 𝑦3+ 𝑥𝑦2+ 𝑦𝑧2+ 𝑥𝑧2+ 𝑥2𝑦 − 4𝑥𝑦𝑧) = 0 => 2𝑧3+ 2𝑥2𝑧 + 2𝑦2𝑧 − 𝑥3− 𝑦3− 𝑥𝑦2− 𝑦𝑧2 − 𝑥𝑧2− 𝑥2𝑦 = 0

4 −𝑥2𝑦

2 +𝑥𝑦2

4 − 2 = 0(∗∗) => 15 (5𝑥3

𝑣ậ𝑦 𝑛ℎâ𝑛 − 3𝑝𝑡(2) − 𝑝𝑡(1) = 0

=> −3(𝑥2− 8𝑥𝑦 + 𝑦2− 8𝑦 + 17𝑥) − (𝑥3+ 3𝑥𝑦2+ 49) = 0

=> −(𝑥 + 1)(𝑥2+ 2𝑥 + 3𝑦2− 24𝑦 + 49) = 0

=> 𝑥2𝑧2(2𝑥

𝑥+𝑧+ 1)2 = (3𝑥2+ 𝑥 + 1) (4𝑥2𝑧4

(𝑥+𝑧) 2) => (3𝑥 + 𝑧)2 = 4𝑧2(3𝑥2+ 𝑥 + 1)

𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖: {9𝑥

2+ 6𝑥𝑧 − 3𝑧2 = 12𝑥2𝑧2+ 4𝑧2𝑥(∗)9𝑧2+ 6𝑥𝑧 − 3𝑥2 = 20𝑥2𝑧2+ 4𝑥2𝑧(∗∗)(đ𝑎𝑛𝑔 𝑥é𝑡 𝑇𝐻: 𝑥 ≠ 0, 𝑦 ≠ 0, 𝑧 ≠0)

𝑧 = 911 𝑣ớ𝑖 𝑥𝑦𝑧 + 2𝑥𝑦 + 2𝑥𝑧 + 2𝑦𝑧 = 0

=> 𝑧(𝑥𝑦 + 2𝑥 + 2𝑦) = −2𝑥𝑦 => 𝑧 = −2𝑥𝑦

𝑥𝑦+2𝑥+2𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1)𝑣à 𝑝𝑡(2): {𝑥

42) { √3 + 2𝑥

2𝑦 − 𝑥4𝑦2+ 𝑥2(1 − 2𝑥2) = 𝑦4(1)

1 + √1 + (𝑥 − 𝑦)2+ 𝑥2(𝑥4− 2𝑥2 − 2𝑥𝑦2+ 1) = 0(2)

ℎệ 𝑡𝑖ế𝑝 𝑡ℎ𝑒𝑜 𝑙à 1 ℎệ đá𝑛ℎ 𝑔𝑖á, 𝑏𝑎𝑛 đầ𝑢 𝑛ℎì𝑛 𝑡ℎì 𝑟ấ𝑡 𝑘ℎó đ𝑜á𝑛 ý 𝑡á𝑐 𝑔𝑖ả 𝑛ℎư𝑛𝑔 𝑡ℎử để ý 𝑥𝑒𝑚 𝑝𝑡(1): √3 + 2𝑥2𝑦 − 𝑥4𝑦2+ 𝑥2− 2𝑥4 = 𝑦4

8 −6(5𝑥

8 )2= 1 => 𝑥 = − 63

340=> 𝑦 = − 63

544 𝑐á𝑐ℎ 2 (𝑇𝑟ươ𝑛𝑔 𝑉ă𝑛 𝐻à𝑜): đặ𝑡 𝑥 = 𝑘𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1), 𝑝𝑡(2)

𝑝𝑡(1): 1

𝑦(2𝑘+1)− (𝑘+1)𝑦

2𝑘 2 𝑦 2 +𝑘𝑦 2 −6𝑦 2 = 1 => −2𝑘−7

=> 𝑘 = −2(𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑘 = −1

2(𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑘 = 8

5(𝑛ℎậ𝑛) 𝑡ℎế 𝑣à𝑜 => 𝑦 = − 63

𝑥𝑖 + 𝑥𝑖(2)

…(𝑥𝑛)3 = (𝑥𝑛)3

(1−𝑛)𝑛(𝑛+1)(𝑛+2)=> 𝑥1 = √ 4

(1−𝑛)𝑛(𝑛+1)(𝑛+2) 3

𝑡ừ đâ𝑦 𝑡𝑎 𝑐ó ℎệ 𝑚ớ𝑖 𝑠𝑎𝑢: { 𝑎

3− 𝑏3 = 8(1)32

(𝑎+𝑏) 6+ 32

(𝑎 3 +𝑏 3 ) 2 = 1(2)

𝑝𝑡(2) 𝑐ó 𝑡ổ𝑛𝑔 𝑠ố 𝑚ũ 𝑙à 6 𝑚à 𝑝𝑡(1)𝑐ó 𝑠ố 𝑚ũ 𝑙à 3 𝑛ê𝑛 𝑐á𝑐ℎ 𝑔𝑖ả𝑖 𝑛𝑔𝑜𝑛 𝑛ℎấ𝑡 𝑙à 𝑠ẽ đẳ𝑛𝑔 𝑐ấ𝑝 𝑛ó 𝑝𝑡(1): 𝑎3− 𝑏3 = 8

=> (𝑎3− 𝑏3)2 = 64 (đ𝑘 𝑎 ≥ 𝑏)

𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): 64

2(𝑎+𝑏) 6+ 64

2(𝑎 3 +𝑏 3 ) 2 = 1 => (𝑎3− 𝑏3)2(𝑎3+ 𝑏3)2+ (𝑎 + 𝑏)6(𝑎3− 𝑏3)2 = 2(𝑎 + 𝑏)6(𝑎3+ 𝑏3)2

𝑣ớ𝑖 𝑥1 = −1 𝑝𝑡(1), 𝑝𝑡(3):

{−1 − 𝑥3 𝑥4 = 2(1)

𝑥3 + 𝑥4 = 2(3) => {

𝑥1 = −1𝑥2 = −1𝑥3 = −1𝑥4 = 3

ℎ𝑎𝑦 {

𝑥1 = −1𝑥2 = −1𝑥3 = 3𝑥4 = −1 𝑣ớ𝑖 1 − 𝑥3 𝑥4 = 0 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1), 𝑝𝑡(3), 𝑝𝑡(4):

𝑥1 𝑥2 + 1 = 2=> 𝑥1 = 𝑥2 = 𝑥3 = 𝑥4 = 1

𝑣ớ𝑖 𝑥4 = −1 => { 𝑥1 + 𝑥2 = 2(1)

−1 − 𝑥1 𝑥2 = 2(3)=> {

𝑥1 = −1𝑥2 = 3𝑥3 = −1𝑥4 = −1

ℎ𝑎𝑦 {

𝑥1 = 3𝑥2 = −1𝑥3 = −1𝑥4 = −1 𝑣ớ𝑖 𝑥1 𝑥2 = 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(3), 𝑝𝑡(4):

{𝑥1 + 𝑥2 = 2

𝑥1 𝑥2 = 1 => {

𝑥1 = 1𝑥2 = 1𝑥3 = 1𝑥4 = 1

47) {

(𝑥 + 𝑦) (1 + 1

𝑥𝑦) = 5(1)(𝑥2+ 𝑦2) (1 + 1

𝑝𝑡(1): 3𝑦 + 6𝑥2 = 9𝑥𝑦(𝑥+√𝑦)

2𝑥 2 +𝑦 => (3𝑦 + 6𝑥2)(2𝑥2+ 𝑦) = 9𝑥2𝑦 + 9𝑥𝑦√𝑦

𝑣ậ𝑦 𝑥 = −5

4−√13

4 ℎ𝑎𝑦 𝑥 =√17

4 −34

=> (2𝑥 − 2𝑦)2− 2(2𝑥 − 2𝑦)√4𝑥2− 2𝑥𝑦 − 2𝑦2+ (√4𝑥2 − 2𝑥𝑦 − 2𝑦2)2 = 0

=> (2𝑥 − 2𝑦 − √4𝑥2− 2𝑥𝑦 − 2𝑦2)2 = 0 => √4𝑥2− 2𝑥𝑦 − 2𝑦2 = 2𝑦 − 2𝑥

𝑡ℎế 𝑣à𝑜 𝑝𝑡(1), 𝑝𝑡(2): {3𝑥

2− 5𝑥𝑦 + 4𝑦2− 2(𝑥 − 𝑦)(2𝑦 − 2𝑥) = 1(∗)2𝑥2− 5𝑥𝑦 + 7𝑦2− 2(𝑥 − 𝑦)(2𝑦 − 2𝑥) = 2(∗∗) 𝑝𝑡(∗): 7𝑥2− 13𝑥𝑦 + 8𝑦2 = 1

𝑝𝑡(∗∗): 6𝑥2− 13𝑥𝑦 + 11𝑦2 = 2

𝑙ấ𝑦 𝑝𝑡(∗)𝑡ℎế 𝑝𝑡(∗∗): 6𝑥2 − 13𝑥𝑦 + 11𝑦2 = 2(7𝑥2− 13𝑥𝑦 + 8𝑦2)

=> 𝑦 = 𝑥 ℎ𝑎𝑦 𝑦 =8𝑥

5 𝑡ℎế 𝑣à𝑜 𝑝𝑡(∗)𝑣ớ𝑖 𝑥 = 𝑦 => 2𝑥2 = 1 => 𝑥 = 𝑦 = ± 1

√2(𝑛ℎậ𝑛) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(∗∗)𝑣ớ𝑖 𝑦 =8𝑥

5 => 𝑥 = ± 5

√167=> 𝑦 = ± 8

√167(𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑡𝑎 𝑐ó 𝑐á𝑐ℎ 𝑘ℎá𝑐 𝑏ị độ𝑛𝑔 ℎơ𝑛 1 𝑡ý 𝑛ℎư𝑛𝑔 𝑣ẫ𝑛 𝑟𝑎 ‼!

để ý 𝑝𝑡(1):𝑥2+𝑥+1

𝑥+1 = 5 +1

𝑦− 𝑥𝑦 => 1

𝑥+1+ 𝑥 = 5 +1

𝑦− 𝑥𝑦 => 𝑥 + 𝑥𝑦 + 1

𝑥+1−1

𝑦 = 5 => 𝑥(1 + 𝑦) +𝑦−𝑥−1

𝑦(𝑥+1)= 5 => 𝑥𝑦 + 𝑥 +𝑦−𝑥−1

𝑦(𝑥+1)= 5 =>𝑦−𝑥−1

𝑦(𝑥+1)= 5 − 𝑥𝑦 − 𝑥

𝑝𝑡(2):𝑦−𝑥−1

𝑦(𝑥+1)= 6

𝑥𝑦+𝑥=> 5 − (𝑥𝑦 + 𝑥) = 6(𝑥𝑦 + 𝑥) => 𝑥𝑦 + 𝑥 =5

7 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1) =>𝑦−𝑥−1

𝑦(𝑥+1)=30

7

𝑡ừ đâ𝑦 𝑡𝑎 𝑐ó ℎệ 𝑚ơ𝑖: {

𝑦−𝑥−1 𝑦(𝑥+1)= 30

2𝑥 2 +𝑦 2 = 4

3𝑥(đ𝑘: 𝑥 ≠ 0) 𝑝𝑡(2): 1 − 1

2𝑥 2 +𝑦 2 = 2

3𝑦(đ𝑘: 𝑦 ≠ 0) 𝑙ấ𝑦 𝑝𝑡(1) + 𝑝𝑡(2) => 2 = 4

3𝑥+ 23𝑦 𝑙ấ𝑦 𝑝𝑡(1) − 𝑝𝑡(2) => 2

2𝑥 2 +𝑦 2 = 4

3𝑥− 23𝑦

𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖: { 18𝑥𝑦 = 12𝑦 + 6𝑥(∗)

18𝑥𝑦 = (2𝑥2+ 𝑦2)(12𝑦 − 6𝑥)(∗∗) 𝑛ℎâ𝑛 (∗), (∗∗): 324𝑥2𝑦2 = (2𝑥2+ 𝑦2)(144𝑦2− 36𝑥2) 𝑣ậ𝑦 𝑦 = 𝑥 ℎ𝑎𝑦 𝑦 = −𝑥

2=> 𝑦 = −1 ℎ𝑎𝑦 𝑦 = 1

2 𝑣ớ𝑖 𝑥 = −1

2 => 𝑦 = −1

2 ℎ𝑎𝑦 𝑦 = 1 𝑣ớ𝑖 𝑥 + 𝑦 + 𝑧 = 0

𝑡𝑎 𝑐ộ𝑛𝑔 3 𝑝𝑡(𝛼), 𝑝𝑡(𝛽), 𝑝𝑡(𝜃) 𝑙ạ𝑖 𝑣ớ𝑖 𝑛ℎ𝑎𝑢

=> 2(𝑥2 + 𝑦2+ 𝑧2) + 𝑥𝑦 + 𝑦𝑧 + 𝑥𝑧 =94

𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑠𝑎𝑢: {(

−3

4 −𝑥𝑧 𝑥+𝑧 )

2+ 𝑧2+ (−

3

4 −𝑥𝑧 𝑥+𝑧 ) 𝑧 =3

4(𝑛)

𝑥2 + 𝑧2+ 𝑥𝑧 =3

4(𝑚) 𝑙ư𝑢 ý 𝑇𝐻 𝑛à𝑦 đ𝑎𝑛𝑔 𝑥 + 𝑧 ≠ 0

16= 0 => (𝑧2−3

2 𝑥 =34

𝑣ớ𝑖 𝑡 =6

7 => √1 + 1

3−𝑥− √1 − 1

3−𝑥 = 67

=> (3 − 𝑥) [9(2 − 𝑥) + 49(3 − 𝑥) + 42√(2 − 𝑥)(3 − 𝑥)] = (4 − 𝑥)[9(3 − 𝑥) +49(2 − 𝑥) + 42√(2 − 𝑥)(3 − 𝑥)

=> (3 − 𝑥)(165 − 58𝑥) − (4 − 𝑥)(125 − 58𝑥) = 42√(2 − 𝑥)(3 − 𝑥)

𝑣ì 𝑡ô𝑖 𝑙à 𝑡á𝑐 𝑔𝑖ả 𝑛ê𝑛 𝑡ô𝑖 𝑠ẽ đư𝑎 𝑠𝑙𝑡 𝑡ℎậ𝑡 𝑠ự 𝑐ủ𝑎 𝑏à𝑖 𝑛à𝑦

để ý 𝑝𝑡(2): 2(𝑥 − 𝑦)2+ 2(𝑥 − 𝑦) + 4(𝑥 + 1) = (𝑥 + 1) + 𝑦 + 2√𝑦(𝑥 + 1) => 2(𝑥 − 𝑦)(𝑥 + 1 − 𝑦) + 3(𝑥 + 1) − 2√𝑦(𝑥 + 1) − 𝑦 = 0

𝑣ậ𝑦 𝑡 = −2 ℎ𝑎𝑦 𝑡 = 1

𝑣ớ𝑖 𝑡 = −2 => 𝑦 = −1

2=> 𝑥 = −2 𝑣ớ𝑖 𝑡 = 1 => 𝑦 = 1 => 𝑥 = 1

59) {

2𝑥 (1 + 1

𝑥 2 +𝑦 2) = 3(1)2𝑦 (1 − 1

𝑥 2 +𝑦 2 = 1

2𝑦 𝑙ấ𝑦 𝑝𝑡(1) + 𝑝𝑡(2): 2 = 3

2𝑥+ 12𝑦 𝑙ấ𝑦 𝑝𝑡(1) − 𝑝𝑡(2): 2

𝑥 2 +𝑦 2 = 3

2𝑥− 12𝑦

𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖 ∶ { 8𝑥𝑦 = 6𝑦 + 2𝑥(∗)

8𝑥𝑦 = (6𝑦 − 2𝑥)(𝑥2+ 𝑦2)(∗∗) 𝑛ℎâ𝑛 𝑝𝑡(∗)𝑣à 𝑝𝑡(∗∗): 64𝑥2𝑦2 = (6𝑦 + 2𝑥)(6𝑦 − 2𝑥)(𝑥2+ 𝑦2)

=> 4𝑥4+ 32𝑥2𝑦2− 36𝑦4 = 0 => 𝑦2 = −𝑥2

9 (𝑙𝑜ạ𝑖 ) ℎ𝑎𝑦 𝑦2 = 𝑥2 𝑣ớ𝑖 𝑦 = 𝑥 𝑡ℎế 𝑣à𝑜 𝑝𝑡(∗): 𝑥 = 0(𝑙𝑜ạ𝑖)ℎ𝑎𝑦 𝑥 = 1 => 𝑦 = 1

4 => 𝑥 =1

4 𝑣ớ𝑖 𝑦 = 4 => 𝑥 = 1

62) { 𝑥2𝑦 + 𝑥2+ 1 = 2𝑥√𝑥2𝑦 + 2(1)

𝑦3(𝑥6− 1) + 3𝑦(𝑥2− 2) + 3𝑦2+ 4 = 0(2)

1−𝑥 2(𝑡𝑎 𝑥é𝑡 𝑥 = ±1 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑛𝑔ℎ𝑖ệ𝑚) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(∗)

𝑣ớ𝑖 𝑥 =√5

2 −1

2=> 𝑦 =1

2+√52 𝑣ớ𝑖 √𝑥2𝑦 + 2 = 𝑥 − 1 𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖 ∶ {𝑥

𝑣ớ𝑖 𝑥 =1

2+√5

2 => 𝑦 =1

2−√52 63) {𝑥

7 − (3 + 𝑦)(𝑥4− 2𝑥2+ 4) − 2𝑥3(𝑦 + 2) + 3𝑥 − 1 = 0(1)

√𝑥2− 𝑦 + 1 + √3(𝑥2 − 𝑦) − 8 = 3(2) 𝑝𝑡(2): đặ𝑡 𝑡 = √𝑥2 − 𝑦 + 1 => 𝑡2− 1 = 𝑥2 − 𝑦(đ𝑘 𝑡 ≥ 0)

=> 𝑡 + √3(𝑡2− 1) − 8 = 3

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

√𝑦−√𝑥−𝑦 3

𝑣ậ𝑦 𝑦 = 0 ℎ𝑎𝑦 𝑦 − 1 − 2√𝑥 − 𝑦 − 𝑥 + 𝑦 + 1 = 1

√𝑦−√𝑥−𝑦 3

𝑣ớ𝑖 𝑦 = 0 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1) => 𝑥 = −1(𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑥 = 5 (𝑑𝑜 đ𝑘 𝑥 ≥ 𝑦)

𝑣ớ𝑖 2𝑦 − 𝑥 − 2√𝑥 − 𝑦 = 1

√𝑦−√𝑥−𝑦 3

=> 2(𝑦 − √𝑥 − 𝑦) − 𝑥 = 1

√𝑦−√𝑥−𝑦 3

𝑝𝑡(1): 𝑥2− 4𝑥 − 5 = 4𝑦2− 4𝑦 − 8𝑦√𝑥 − 𝑦

=> 𝑥2− 5 = 4(√𝑥 − 𝑦 − 𝑦)2

𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖: {𝑥

2− 5 = 4𝑡2(∗)2𝑡 − 𝑥 = 1

=> 𝑎2

√2+𝑎+ 𝑏2

√2+𝑏= √2 => 𝑎2(√2 + 𝑏) + 𝑏2(√2 + 𝑎) = √2(√2 + 𝑎)(√2 + 𝑏) => (𝑎2+ 𝑏2)√2 + 𝑎2𝑏 + 𝑏2𝑎 = 2√2 + 2𝑎 + 2𝑏 + 𝑎𝑏√2(∗∗)

𝑣ớ𝑖 𝑎 + 𝑏 = √2 𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖: {𝑎 + 𝑏 = √2(𝜀)

𝑎2+ 𝑏2 = 4(∗)

𝑑𝑒𝑛𝑡𝑎 = −(64𝑥4+ 32𝑥3+ 108𝑥2+ 20𝑥 + 31) < 0 (𝑣ớ𝑖 𝑥 ≥ 1

√4

3 ) 𝑣ậ𝑦 𝑝𝑡(𝛿)𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚

𝑣ớ𝑖 𝑡 = 2 => 𝑥 =3

2 𝑙ạ𝑖 1 𝑙ầ𝑛 𝑛ữ𝑎 ℎ𝑝𝑡 𝑙ạ𝑖 𝑡ℎể ℎ𝑖ệ𝑛 đượ𝑐 𝑠ứ𝑐 𝑚ạ𝑛ℎ 𝑐ủ𝑎 𝑚ì𝑛ℎ 𝑡𝑟ê𝑛 𝑛𝑔ườ𝑖 𝑎𝑛ℎ 𝑒𝑚 𝑐ủ𝑎 𝑛ó 𝑙à 𝑝𝑡

𝑝𝑡(1): 2𝑦2+ 2𝑦 − 84 = 17 √49 − (𝑥 + 𝑦)6 2

𝑡𝑎 𝑙𝑢ô𝑛 𝑐ó 17 √49 − (𝑥 + 𝑦)4 2

≥ 0 => 2𝑦2 + 2𝑦 − 84 ≥ 0 => 𝑦 ≥ 6 ℎ𝑎𝑦 𝑦 ≤ −7

𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖 {(−1 − 𝑦)(−1 + 𝑧) = 2 − 𝑦𝑧(𝑎)

𝑦(−1 + 𝑧) = 3 − (−1 − 𝑦)𝑧(𝑏)

=> { 𝑦 − 𝑧 = 1(𝑎)

𝑦 + 𝑧 = −3(𝑏)=> {

𝑦 = −1

𝑧 = −2 => 𝑥 = 0 𝑣ớ𝑖 𝑥 + 𝑧 = 0 𝑣à 𝑦 + 𝑧 = 0 𝑡ℎì ℎ𝑝𝑡 𝑛à𝑦 𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚

√𝑥−𝑦+1+ 1

√𝑥−1+√𝑦+ 1 = 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 ≥ 1) 𝑣ớ𝑖 𝑥 − 𝑦 − 1 = 0 => 𝑦 = 𝑥 − 1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1):

đặ𝑡 𝑡 = 4𝑥3− 𝑥 + 3 => 2𝑡3 − 2𝑥3 = 3

𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑠𝑎𝑢: { 2𝑡

3− 2𝑥3 = 3(∗)

𝑡 = 4𝑥3− 𝑥 + 3(∗∗) 𝑙ấ𝑦 𝑝𝑡(∗)𝑡ℎế 𝑣à𝑜 𝑝𝑡(∗∗):

=> 𝑡 = 4𝑥3− 𝑥 + 2𝑡3− 2𝑥3 = 2𝑥3 + 2𝑡3− 𝑥

=> 𝑡 + 𝑥 = 0 ℎ𝑎𝑦 2𝑡2 − 2𝑥𝑡 + 2𝑥2 = 1

𝑣ớ𝑖 𝑡 + 𝑥 = 0 => 4𝑥3 − 𝑥 + 3 + 𝑥 = 0 => 𝑥 = √3

4 3

𝑦= 4 => 1

𝑦= 4 − 𝑥 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1):

𝑛ế𝑢 𝑥 ≠ −1 => 𝑦 + √𝑥𝑦 =𝑥+14 + 𝑥2 − 𝑥 𝑡ℎế 𝑣à𝑜 𝑝𝑡(∗)

=> 1

√𝑥+√𝑦+

4 𝑥+1 +𝑥2−𝑥−2

√𝑥𝑦+(𝑥−𝑦)√𝑥𝑦−2)+𝑦= 0(∗∗) 𝑏â𝑦 𝑔𝑖ờ 𝑡𝑎 𝑐ℎỉ 𝑐ầ𝑛 𝑐ℎứ𝑛𝑔 𝑚𝑖𝑛ℎ 4

𝑥+1+ 𝑥2− 𝑥 − 2 ≥ 0 ℎ𝑜ặ𝑐 > 0 𝑛ữ𝑎 𝑙à 𝑥𝑜𝑛𝑔 ‼! đặ𝑡 𝑓(𝑥) = 4

=> 𝑥4+ 2𝑥2+ 1 − 2𝑥2− 4𝑥 − 2 = 0

=> (𝑥2 + 1)2− 2(𝑥 + 1)2 = 0

𝑣ậ𝑦 𝑥2+ 1 − √2(𝑥 + 1) = 0 ℎ𝑎𝑦 𝑥2+ 1 + √2(𝑥 + 1) = 0

𝑣ớ𝑖 𝑥2+ 1 − √2(𝑥 + 1) = 0 => 𝑥 =1±√2√2−1

√2 𝑣ớ𝑖 𝑥2+ 1 + √2(𝑥 + 1) = 0 => 𝑝𝑡𝑣𝑛

=> 𝑥2+ 𝑥 + 1 = 𝑥4− 𝑥2+ 2𝑥3 − 2𝑥 + 1

=> 𝑥4+ 2𝑥3− 2𝑥2− 3𝑥 = 0 => 𝑥 = 0(𝑙𝑜ạ𝑖)ℎ𝑎𝑦 𝑥 = −1(𝑙𝑜ạ𝑖)ℎ𝑎𝑦 𝑥 = −1±√13

2 𝑣ớ𝑖 √𝑥2+ 𝑥 + 1 = 3 − 𝑥 => 𝑥2 + 𝑥 + 1 = (3 − 𝑥)2(đ𝑘: 3 − 𝑥 ≥ 0 => 𝑥 ≤ 3)

2 =3𝑎2𝑏2

2 + 𝑏4

=> 𝑏(𝑎 − 𝑏)(𝑎2+ 𝑎𝑏 + 𝑏2) +3𝑎𝑏2

2 (𝑏 − 𝑎) = 0 𝑣ậ𝑦 𝑎 = 𝑏 ℎ𝑎𝑦 𝑏(𝑎2+ 𝑎𝑏 + 𝑏2) −3𝑎𝑏2

2 = 0(∗) 𝑣ớ𝑖 𝑝𝑡(∗):1

2𝑏(2𝑎2− 𝑎𝑏 + 2𝑏2) = 0

𝑣ậ𝑦 𝑏 = 0(𝑙𝑜ạ𝑖) ℎ𝑎𝑦 2𝑎2− 𝑎𝑏 + 2𝑏2 = 0

𝑣ớ𝑖 2𝑎2− 𝑎𝑏 + 2𝑏2 = 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑎, 𝑏 = 0 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎệ) 𝑣ớ𝑖 𝑎 = 𝑏 => √𝑥 + 1 = √2𝑦 => 𝑦 =𝑥+12 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2):

2 + 𝑡𝑘 +3𝑘2

2 = 𝑡 + 𝑘 => −1

𝑎𝑏 = 18(∗∗) => {

𝑡𝑎𝑛𝑦 = −9

2𝑡𝑎𝑛𝑧 = −4 ℎ𝑎𝑦 {

𝑡𝑎𝑛𝑦 = 9

2𝑡𝑎𝑛𝑧 = 4 𝑣ớ𝑖 𝑡𝑎𝑛𝑧 = −4 => 𝑡𝑎𝑛𝑥 = −1

2 𝑣ớ𝑖 𝑡𝑎𝑛𝑧 = 4 => 𝑡𝑎𝑛𝑥 =1

2

3𝑥 2 +4𝑥−3 + 𝑥 = 0

2 ) = 𝑧(𝑥 + 1) => 𝑧 = − 𝑥2

5𝑥+2 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2):

=> 𝑦 + 1 =−𝑥+

3𝑥2 5𝑥+2

2 => 𝑦 =−𝑥2−6𝑥−2

5𝑥+2 𝑡ừ đâ𝑦 𝑡ℎế 𝑛𝑔ượ𝑐 𝑣à𝑜 𝑝𝑡(1): => (−𝑥2−6𝑥−2

3 => 𝑦 = −1

3=> 𝑧 = −1

3 𝑣ớ𝑖 𝑥 = 2 => 𝑦 = −3

3 ≤ 1

3(17−2𝑥+9

2 ) =26−2𝑥

6 => 𝑉𝑇 ≤ 0 𝑚à 𝑐ũ𝑛𝑔 𝑐ó 𝑉𝑇 ≥ 0 𝑣ậ𝑦 𝑑ấ𝑢 𝑏ằ𝑛𝑔 𝑥ả𝑦 𝑟𝑎 𝑘ℎ𝑖 𝑥 = 4

=> 𝑥 + 3𝑥−

1 2𝑥−3

𝑥 2 +(2𝑥−3)21 = 3 (đ𝑘: 𝑥 =3

2 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑛ê𝑛 𝑥 ≠3

2) 𝑣ậ𝑦 𝑥 = 1 ℎ𝑎𝑦 𝑥 = 0(𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑥 = 2

=> 𝑦 = 1

1−𝑥 (đ𝑘: 𝑥 ≠ 1) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1):

=> 𝑥 + 𝑥+

2 1−𝑥

𝑥 2 + 1

(1−𝑥)2

= 2 => 𝑥 = 0 ℎ𝑎𝑦 𝑥 = 2

𝑡ạ𝑖 𝑠𝑎𝑜 𝑡ô𝑖 𝑙ạ𝑖 𝑙ấ𝑦 2 𝑏à𝑖 𝑛à𝑦 để 𝑔𝑖ả𝑖 𝑏ở𝑖 𝑣ì 𝑡ừ 𝑡𝑟ướ𝑐 𝑘ℎ𝑖 𝑛ℎắ𝑐 đế𝑛 𝑏à𝑖 𝑛à𝑦 𝑡ℎì đề𝑢 𝑥à𝑖 𝑝ℎứ𝑐 ℎó𝑎 𝑣à 𝑡ô𝑖 𝑙ạ𝑖 đặ𝑡 𝑣ấ𝑛 đề 𝑛ế𝑢 𝑘ℎô𝑛𝑔 𝑥à𝑖 𝑝ℎứ𝑐 ℎó𝑎 𝑐ó đượ𝑐 𝑘ℎô𝑛𝑔 ? ? ? 𝑣à đâ𝑦 𝑙à 𝑘ế𝑡 𝑞𝑢ả

95) {𝑥

3 − 3𝑥2 + 2 = √𝑦3+ 3𝑦2(1)

3√𝑥 − 2 = √𝑦2+ 8𝑦(2)

𝑥) = 𝑥4+ 14𝑥2+ 49 => (4𝑥2 + 8𝑥 + 4) (𝑥 +3

𝑥) = 𝑥4+ 14𝑥2+ 49 => (4𝑥2 + 8𝑥 + 4)(𝑥2+ 3) = 𝑥5+ 14𝑥3+ 49𝑥

=> 4𝑥4+ 12𝑥2+ 8𝑥3+ 24𝑥 + 4𝑥2+ 12 = 𝑥5+ 14𝑥3 + 49𝑥

=> 𝑥5− 4𝑥4+ 6𝑥3− 16𝑥2+ 25𝑥 − 12 = 0

=> (𝑥 − 1)2(𝑥 − 3)(𝑥2+ 𝑥 + 4) = 0

𝑡ừ đâ𝑦 𝑡𝑎 𝑡ℎ𝑢 đượ𝑐 đ𝑘 𝑙à ∶ 𝑥, 𝑦, 𝑧 > 0

𝑡ℎ𝑒𝑜 𝑐𝑎𝑢𝑐ℎ𝑦 𝑡𝑎 𝑐ó ∶ {

𝑦(𝑥2+ 1) ≥ 2𝑦√𝑥2 = 2𝑦𝑥(∗)𝑧(𝑦4+ 𝑦2+ 1) ≥ 3𝑧 √𝑦3 6

= 3𝑧𝑦2(∗∗)𝑥(𝑧6+ 𝑧4+ 𝑧2+ 1 ≥ 4𝑥 √𝑧4 12= 4𝑥𝑧3(∗∗∗) 𝑛ℎâ𝑛 3 𝑝𝑡(∗), 𝑝𝑡(∗∗), 𝑝𝑡(∗∗∗) 𝑛ê𝑛 𝑡𝑎 đượ𝑐 𝑉𝑃 ≥ 24𝑥2𝑦3𝑧4 = 𝑉𝑇

𝑑ấ𝑢 𝑏ằ𝑛𝑔 𝑐ℎỉ 𝑥ả𝑦 𝑟𝑎 𝑘ℎ𝑖 ∶ 𝑥 = 𝑦 = 𝑧 = 1

101) {√𝑥 + √𝑦 + 2√2 + 𝑥 + 𝑦 + (√𝑥 − √𝑦)

2

= 6(1)4𝑥√𝑥2 + 3 + 9𝑦√𝑦2+ 8 = 5𝑥2+ 15𝑦2+ 15(2)

để ý 𝑝𝑡(2)𝑙à 1 𝑑ạ𝑛𝑔 𝑆 𝑂 𝑆 𝑛ℎư𝑛𝑔 𝑣ấ𝑛 đề ở đâ𝑦 𝑙à 𝑡á𝑐ℎ 𝑙à𝑚 𝑠𝑎𝑜 để 𝑥𝑢ấ𝑡 ℎ𝑖ệ𝑛 𝑛ó ? ? => 5𝑥2+ 15𝑦2 + 15 − 4𝑥√𝑥2+ 3 − 9𝑦√𝑦2+ 8 = 0

𝑛=> −2

𝑛(1 + 𝑛2) = 5 => −2 − 2𝑛2 = 5𝑛 => 𝑛 = −2 ℎ𝑎𝑦 𝑛 = −1

2

𝑛 = −2 => 𝑡 = 1 ℎ𝑎𝑦 𝑣ớ𝑖 𝑛 = −1

2=> 𝑡 = 4 𝑙ấ𝑦 𝑝𝑡(𝑎) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(𝑏): 𝑘 = −9

2𝑙 => −9

2𝑙(1 + 𝑙2) = 15 => −9 − 9𝑙2 = 30𝑙 => 𝑙 = −1

3 ℎ𝑎𝑦 𝑙 = −3 𝑣ớ𝑖 𝑙 = −1

3=> 𝑘 =27

2 ℎ𝑎𝑦 𝑣ớ𝑖 𝑙 = −3 => 𝑘 =3

2

𝑡ừ đâ𝑦 𝑡𝑎 𝑐ó 𝑐á𝑐 ℎệ 𝑠ố 𝑝ℎù ℎợ𝑝 ℎệ 𝑠𝑎𝑢 𝑙à ∶

{

𝑛 = −

1 2

𝑡 = 4

𝑙 = −13

𝑘 =272

𝑡ừ đâ𝑦 𝑙ắ𝑝 𝑣à𝑜 𝑝𝑡 𝑡𝑟ê𝑛:

=> 4 (𝑥 −√𝑥2+3

2 )

2+27

102) 𝑥 = √2 + √2 − √2 + 𝑥

đâ𝑦 𝑙à 𝑠𝑙𝑡 𝑐ủ𝑎 𝑡ℎầ𝑦 𝑂𝑟𝑙𝑎𝑛𝑑𝑜 𝐼𝑟𝑎ℎ𝑜𝑙𝑎 𝑂𝑟𝑡𝑒𝑔𝑎 𝑏ằ𝑛𝑔 𝑐á𝑐ℎ 𝑙ượ𝑛𝑔 𝑔𝑖á𝑐 𝑐ự𝑐 đẹ𝑝 ‼! đặ𝑡 𝑥 = 2𝑐𝑜𝑠𝑡(đ𝑘: − 1 ≤ 𝑥 ≤ 1)

=> 4cos2𝑡 = 2 + 2 sin (𝑡

4) => 𝑐𝑜𝑠2𝑡 = sin (𝑡

4) => sin (𝑝𝑖

2 + 2𝑡) = sin (𝑡

4) => 𝑥 = 2 cos (2

9𝑝𝑖)