\(\left(x^2+x\right)^2+4x^2+4x-12=\left[\left(x^2+x\right)^2+4\left(x^2+x\right)+4\right]-16=\left(x^2+x+2\right)-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

\(\left(x^2+x\right)^2+4x^2+4x-12\\ =\left(x^2+x+2\right)-4\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\)

\(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)

\(=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)

A. x² - 3xy

=  x (x - 3y)

B. (x + 5)² - 9

=  (x + 5) - 3²

=  (x + 5 + 3) (x + 5 - 3)

= ( x + 8) ( x + 2)

C. xy + xz - 2y - 2z

= (xy + xz) - (2y + 2z)

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

= (x - 2) (y + z)

làm ơn trả lời câu hỏi của mình đi

a) \(x^2-5xy+6y^2\)

\(=x^2-3xy-2xy+6y^2\)

\(=x\left(x-3y\right)-2y\left(x-3y\right)\)

\(=\left(x-2y\right)\left(x-3y\right)\)

b) \(16\left(x-1\right)^2-36y^2\)

\(=\left(4x-4\right)^2-\left(6y\right)^2\)

\(=\left(4x+6y-4\right)\left(4x-6y-4\right)\)

c) \(4\left(x+y\right)-12\left(x+y\right)^2\)

\(=\left(x+y\right)\left[4-12\left(x+y\right)\right]\)

\(=4\left(x+y\right)\left[1-3x-3y\right]\)

Đa thức này không phân tích được thành nhân tử.

Bạn coi lại đề.

Ta có: \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)+3x^2\)

\(=4\left(x^2+60+17x\right)\left(x^2+60x+16x\right)+3x^2\)

\(=4\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]+3x^2\)

\(=4\left(x^2+60\right)^2+132x\left(x^2+60\right)+1091x^2\)

\(A=\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(A=\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2\)

\(A=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

Đặt \(p=x^2-4,5x-8\)ta có :

\(A=\left(p-2,5x\right)\left(p+2,5x\right)+4x^2\)

\(A=p^2-\left(2,5x\right)^2+4x^2\)

\(A=p^2-6,25x^2+4x^2\)

\(A=p^2-2,25x^2\)

\(A=p^2-\left(1,5x\right)^2\)

\(A=\left(p-1,5x\right)\left(p+1,5x\right)\)

Thay \(p=x^2-4,5x-8\)vào A ta có :

\(A=\left(x^2-4,5x-8-1,5x\right)\left(x^2-4,5x-8+1,5x\right)\)

\(A=\left(x^2-6x-8\right)\left(x^2-3x-8\right)\)

\(\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(=\left(x+1\right)\left(x-8\right)\left(x-4\right)\left(x+2\right)+4x^2\)

\(=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

  Đặt \(x^2-2x-8=t\)

  Ta có : \(\left(t-5x\right)t+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-2.\frac{5}{2}xt+\frac{25}{4}x^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}\right)^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}-\frac{3}{2}x\right)\left(t-\frac{5}{2}+\frac{3}{2}x\right)\)

    Học tốt ~~

\(=x^2\left(x^2+2x+1\right)+x+1\)

\(=x^2\left(x+1\right)^2+x+1\)

\(=\left(x+1\right)\left[x^2\left(x+1\right)+1\right]\)

\(=\left(x+1\right)\left(x^3+x^2+1\right)\)

\(x^4+2x^3+x^2+x+1\)

\(=x^2\left(x+1\right)^2+\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3+x^2+1\right)\)

\(=4ab\left(ab+ax+bx+x^2\right)=4a^2b^2+4a^2bx+4ab^2x+4abx^2\)

4ab(a + x)(b + x) là nhân tử rồi nhé