Question

Cho f(x) = (x+2)(x+3)(x+4)(x+5) + 1. CMR f(x) luôn có giá trị chính phương với mọi x nguyên

Answer 1

\(f\left(x\right)=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+1\)  

\(f\left(x\right)=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)+1\)

\(f\left(x\right)=\left(x^2+5x+2x+10\right)\left(x^2+4x+3x+12\right)+1\)

\(f\left(x\right)=\left(x^2+7x+10\right)\left(x^2+7x+12\right)+1\)

\(f\left(x\right)=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)+1\)

\(f\left(x\right)=\left(x^2+7x+11\right)^2-1+1\)

\(f\left(x\right)=\left(x^2+7x+11\right)^2\Leftrightarrowđpcm\)

Answer 2

ƒ (x)=(x+2)(x+3)(x+4)(x+5)+1  

ƒ (x)=(x+2)(x+5)(x+3)(x+4)+1

ƒ (x)=(x2+5x+2x+10)(x2+4x+3x+12)+1

ƒ (x)=(x2+7x+10)(x2+7x+12)+1

ƒ (x)=(x2+7x+11−1)(x2+7x+11+1)+1

ƒ (x)=(x2+7x+11)2−1+1

ƒ (x)=(x2+7x+11)2⇔đpcm