Question

phân tích thành nhân tử

A=(x+1)(x+2)(x+3)(x+4)-8

Answer 1

\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-8\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-8\)

Đặt \(x^2+5x+5=t\)

Khi đó: \(A=\left(t-1\right)\left(t+1\right)-8\)

              \(=t^2-9=\left(t-3\right)\left(t+3\right)\)

              \(=\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)

Chúc bạn học tốt.

Answer 2

A=\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-8\)

A=\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-8\)

A=\(\left(x^2+5x +4\right)\left(x^2+5x+6\right)-8\)

Đặt \(x^2+5x+4=x\)ta có:

x(x+2)-8=\(x^2+2x-8\)=\(\left(x+1\right)^2-9\)=(x+1-3)(x+1+3)=(x-2)(x+4)=\(\left(x^2+5x+4-2\right)\left(x^2+5x+4+4\right)\)=\(\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)