Đáp án C

G T ⇔ x 2 + y − 3 x + y 2 − 4 y + 4 = 0 y 2 + x − 4 y + x 2 − 3 x + 4 = 0

có nghiệm  ⇔ Δ x ≥ 0 Δ y ≥ 0 ⇔ 0 ≤ x ≤ 4 3 1 ≤ y ≤ 7 3

Và:

x y = 3 x + 4 y − x 2 − y 2 − 4 ⇒ P = 3 x 3 + 18 x 2 + 45 x − 8 ⏟ f x + − 3 y 3 + 3 y 2 + 8 y ⏟ g y

 Xét hàm số f x = 3 x 3 + 18 x 2 + 45 x − 8 trên  0 ; 4 3 ⇒ max 0 ; 4 3 f x = f 4 3 = 820 9

Xét hàm số g x = − 3 y 3 + 3 y 2 + 8 y trên  1 ; 7 3 ⇒ max 1 ; 7 3 g x = f 4 3 = 80 9

Vật P ≤ max 0 ; 4 3 f x + max 1 ; 7 3 g x = 100

Dấu “=” xảy ra khi  x = y = 4 3

6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)

\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)

7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)

\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)

8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)

\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)

9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)

10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)

\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)

11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)

12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)

13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)

a) \(\left(x^5+4x^3-6x^2\right):4x^2\)

\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)

\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)

b) 

Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)

c) (-2x⁵ : 2x²) + (3x² : 2x²) + (-4x^3 : 2x^2)

= \(-x^3+\dfrac{3}{2}-2x\)

d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)

\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)

\(=x-4\)

(dùng hẳng đẳng thức thứ 7)

Bài 2 :

a) 3x(x - 2) - 5x(1 - x) - 8(x² - 3)

= 3x² - 6x - 5x + 5x² - 8x² + 24

= (3x² + 5x² - 8x²) + (-6x - 5x) + 24 

= -11x + 24

b) (x - y)(x² + xy + y²) + 2y³

= x³ - y³ + 2y³

= x³ + y³ 

c) (x - y)² + (x + y)² - 2(x - y)(x + y)

= (x - y)² - 2(x - y)(x + y) + (x + y)²

= [(x - y) + x + y)² = [x - y + x + y] = (2x)² = 4x²

Bài 1 :

a]=  \(\frac{1}{4}\)x³ + x - \(\frac{3}{2}\).

b] => [x³ + x² -12 ] = [ x² +3 ][x-2] + [-6]

c]= -x³ -2x +\(\frac{3}{2}\).

d] = [ x³ - 64 ]  = [ x² + 4x + 16][ x- 4].

1. Ta có: hằng đẳng thức: \(x^3+y^3+z^3=3xyz\) nếu x+y+z=0

đặt b-c=x, c-a=y, a-b=z⇒x+y+z=0

 \(\Rightarrow\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3=3\left(a-b\right)\left(c-a\right)\left(b-c\right)\)

2. \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

3. Tham khảo: https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-y-5-x-5-y-5-thanh-nhan-tu-faq447273.html

\(5,=x^3+2x^2y-7x^2y-14xy^2\\ =x^2\left(x+2y\right)-7xy\left(x+2y\right)\\ =x\left(x-7y\right)\left(x+2y\right)\)

Đáp án đúng : A

Mình xp giúp được mỗi câu đầu thôi nha ;-;;;; 2 câu sau mình chưa học, bạn thông cảm ;-;;;.

`a,` \(\text{P(x) =}\)\(2x^3-3x+x^5-4x^3+4x-x^5+x^2-2\)

`P(x)= (2x^3 - 4x^3)-(3x-4x) +(x^5-x^5) +x^2-2`

`P(x)= -2x^3- (-x)+0+x^2-2`

`P(x)=-2x^3+x+x^2-2`

`Q(x)= x^3-x^2+3x+1+3x^2`

`Q(x)= x^3- (x^2-3x^2) +3x+1`

`Q(x)=x^3- (-2x^2)+3x+1`

\(a,=3xy^2\\ b,=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\\ c,=-2x^2+xy+5x^3y^2\\ d,=\left(3x-y\right)\left(9x^2+3xy+y^2\right):\left(3x-y\right)=9x^2+3xy+y^2\)

\(\left(5x^5y^2z+\dfrac{1}{2}x^4y^2z^3-2xy^3z^2\right):\dfrac{1}{4}xy^2z\\ =\left(5:\dfrac{1}{4}\right).\left(x^5:x\right).\left(y^2:y^2\right).\left(z:z\right)+\left(\dfrac{1}{2}:\dfrac{1}{4}\right).\left(x^4:x\right).\left(y^2:y^2\right).\left(z^3:z\right)-\left(2:\dfrac{1}{4}\right).\left(x:x\right).\left(y^3:y^2\right).\left(z^2:z\right)\\ =20x^4+2x^3z^2-8yz\)