tick cho mình rồi mình làm cho

a. \(\left(x^2+2x\right)^2+9x^2+18x+20=x^4+4x^3+13x^2+18x+20\)

\(=x^4+2x^3+2x^3+5x^2+4x^2+4x^2+8x+10x+20\)

\(=x^2\left(x^2+2x+5\right)+2x\left(x^2+2x+5\right)+4\left(x^2+2x+5\right)=\left(x^2+2x+5\right)\left(x^2+2x+4\right)\)

Lưu ý: có thể dùng phương pháp đồng nhất hệ số dưới dạng \(\left(x^2+ax+5\right)\left(x^2+bx+4\right)\) khi thực xong bước 1

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

c. \(x^2-4xy+4y^2-2x+4y-35=\left(x-2y\right)^2-2\left(x-2y\right)+1-36=\left(x-2y-1\right)^2-6^2\)

\(=\left(x-2y-1-6\right)\left(x-2y-1+6\right)=\left(x-2y-7\right)\left(x-2y+5\right)\)

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

x² - 2x + 2y - xy = (x² - 2x) - (xy - 2y) = x(x - 2) - y(x - 2) = (x - 2)(x - y)

x² + 4xy - 16 + 4y² = (x² + 4xy + 4y²) - 16 = (x + 2y)² - 16 = (x + 2y + 4)(x + 2y - 4)

cảm ơn nha

\(\left(x+2y-4\right)\left(x+2y+4\right)\)

(x+y)^2 - 5y^2

x^2- 4y^2 + 4xy

= x^2 + 4xy - 4y^2

=x^2 + 2x2y - (2y)^2

= ( x - 2y )^2

 x² - 4y² + 4xy = x² + 4xy - 4y² = - (x² - 4xy + 4y²) = - ((x² - 2x.2y + (2y)²)

= - ( x - 2y)²

\(x^2-25-4xy+4y^2\)

\(=\left(x^2-4xy+4y^2\right)-25\)

\(=\left[x^2-2\cdot x\cdot2y+\left(2y\right)^2\right]-25\)

\(=\left(x-2y\right)^2-5^2\)

\(=\left(x-2y-5\right)\cdot\left(x-2y+5\right)\)