\(\left(x^2+2x\right)^2+4\left(x^2+2x\right)+5\left(x^2+2x\right)+20\)

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

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

(x2+2x)+9x2+18x+20

=(x2+2x)+9(x2+2x)+20

Đặt t=x2+2x đc:

t+9t+20=10t+20=10(t+2)

Thay t=x2+2x vào đc:

10(x2+2x+2)

\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)

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

(x²+2x)²+9x²+18x+20

=(x²+2x)²+9(x²+2x)+20

Đặt t=x²+2x ta được:

t²+9t+20=t²+4t+5t+20

=t.(t+4)+5.(t+4)

=(t+4)(t+5)

thay t=x²+2x ta được:

(x²+2x+4)(x²+2x+5)

Vậy (x²+2x)²+9x²+18x+20=(x²+2x+4)(x²+2x+5)

(x² + 2x + 4) (x² + 2x + 5)

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

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

      \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

\(=\left(x^2+x\right)^2-5\left(x^2+x\right)+3\left(x^2+x\right)-15\)

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

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

      \(\left(x^2+2x\right)^2+9x^2+18x+20\)

\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)

\(=\left(x^2+2x\right)^2+5\left(x^2+2x\right)+4\left(x^2+2x\right)+20\)

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

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

a, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

Gọi \(x^2+x=A\)

\(\Rightarrow A^2-2A-15\)

\(\Rightarrow\left(A-3\right)\left(A+5\right)\)

\(\Rightarrow\left(x^2+x-3\right)\left(x^2+x+5\right)\)

a) ( 4x+1) (12x-1) (3x+2) (x+1) -4

=(4x+1)(3x+2)(12x-1)(x+1)-4

=(12x²+11x+2)(12x²+11x-1)-4

Đặt t=12x²+11x+2 ta được:

t.(t-3)-4

=t²-3t-4

=t²+t-4t-4

=t.(t+1)-4.(t+1)

=(t+1)(t-4)

thay t=12x²+11x+2 ta được:

(12x²+11x+3)(12x²+11x-2)

Vậy ( 4x+1) (12x-1) (3x+2) (x+1) -4=(12x²+11x+3)(12x²+11x-2)

b) (x²+2x)²+9x²+18x+20

=(x²+2x)²+9.(x²+2x)+20

Đặt y=x²+2x ta được:

y²+9y+20

=y²+4y+5y+20

=y.(y+4)+5.(y+4)

=(y+4)(y+5)

thay y=x²+2x ta được:

(x²+2x+4)(x²+2x+5)

Vậy (x²+2x)²+9x²+18x+20=(x²+2x+4)(x²+2x+5)