Question

( x + 1 )(x+2)(x+3)(x+4)-24.PhÂn tích thành nhân tư phuong phap (a+x)(b+x)(c+x)(d+x) vƠI (a+b)=(c+d)

CẢM ơn ạ

Answer 1

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)   (1)

Đặt \(t=x^2+5x+4\)

(1) \(=t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=t^2+2t+1-25\)

\(=\left(t+1\right)^2-5^2\)

\(=\left(t+1-5\right)\left(t+1+5\right)\)

\(=\left(t-4\right)\left(t+6\right)\)

\(\Rightarrow\left(1\right)=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)

Answer 2

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=x\left(x+2\right)\left(x+3\right)\left(x+4\right)+\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=x^2\left(x+3\right)\left(x+4\right)+2x\left(x+3\right)\left(x+4\right)+x\left(x+3\right)\left(x+4\right)+2\left(x+3\right)\left(x+4\right)-24\)

\(=x^2\left(x+3\right)\left(x+4\right)+3x\left(x+3\right)\left(x+4\right)+2\left(x+3\right)\left(x+4\right)-24\)

\(=x^3\left(x+4\right)+3x^2\left(x+4\right)+3x^2\left(x+4\right)+9x\left(x+4\right)+2x\left(x+4\right)+6\left(x+4\right)-24\)

\(=x^3\left(x+4\right)+6x^2\left(x+4\right)+11x\left(x+4\right)+6\left(x+4\right)-24\)

\(=x^4+4x^3+6x^3+24x^2+11x^2+44x+6x+24-24\)

\(=x^4+10x^3+35x^2+50x\)

\(=x\left(x+5\right)\left(x^2+5x+10\right)\)