Question

CMR:

A=(x+1)(x+2)(x+3)(x+4)-24 Chia hết cho (x+5) với mọi x khác 5

Answer 1

Ta có: A = (x + 1)(x + 2)(x + 3)(x + 4) - 24

A = (x + 1)(x + 4)(x + 2)(x + 3) - 24

A = (x² + 5x + 4)(x² + 5x + 6) - 24

Đặt x² + 5x + 4 = k

=> k(k + 2) - 24 = k² + 2k - 24 = k² + 6k - 4k - 24 = k(k + 6) - 4(k + 6) = (k - 4)(k + 6)

      => (x² + 5x + 4 - 4)(x² + 5x +  4 + 6) = (x² + 5x)(x² + 5x + 10) = x(x  + 5)(x² + 5x + 10)

Do x + 5 \(⋮\)x + 5 => x(x + 5)(x² + 5x + 10) \(⋮\)x + 5

Answer 2

thế cần CM cho x khác 5 ko

Answer 3

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

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

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

Đặt \(x^2+5x+5=a\)

\(\Rightarrow A=\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)

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