Question

Chứng minh rằng với mọi x thuộc N thì M= (x+1)(x+3)(x+5)(x+7)+15 chia hết cho x+6

Answer 1

\(M=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

\(=\left(x^2+8x+11\right)^2-16+15=\left(x^2+8x+11\right)^2-1=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

\(\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)⋮\left(x+6\right)\)

Answer 2

\(M=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(\Rightarrow M=x^4+16x^3+86x^2+176x+120\)

\(\Rightarrow M=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

\(\Rightarrow M=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)

Sau khi phân tích đa thức M thành nhân tử, ta thấy: M chứa thừa số x + 6 nên \(M⋮\left(x+6\right)\)

Vậy với mọi \(x\inℕ\)thì\(M=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15⋮\left(x+6\right)\)

Answer 3

\(M=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+6\right)+\left(x+1\right)\left(x+3\right)\left(x+6\right)-\left(x+1\right)\left(x+6\right)+\)

\(\left(x+1\right).3+15\)

\(=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+6\right)+\left(x+1\right)\left(x+3\right)\left(x+6\right)-\left(x+1\right)\left(x+6\right)+3\left(x+6\right)\)

\(=\left(x+6\right)\left[\left(x+1\right)\left(x+3\right)\left(x+5\right)+\left(x+1\right)\left(x+3\right)-\left(x+1\right)+3\right]\)chia hết cho x+6