Question

Cho x thuộc tập hợp Z, chứng minh rằng A thuộc tập hợp Z. Biết:

A = \(\sqrt{x^4+2x^3+3x^2+2x+1}\)

Answer 1

`A = sqrt(x^4 + 2x^3 + 3x^2 + 2x + 1)`

`A = sqrt(x^4 + x^3 + x^2 + x^3 + x^2 + x + x^2 + x + 1)`

`A = sqrt(x^2(x^2+x+1) + x(x^2+x+1) + 1(x^2+x+1))`

`A = sqrt((x^2+x+1)(x^2+x+1))`

`A = sqrt((x^2+x+1))^2`

`A = x^2 + x + 1 ( x^2 + x + 1 = x^2 + 1/2x + 1/2x + 1/4 + 3/4 = (x+1/2)^2 +3/4 >0 forall x`.

Vì `x in ZZ -> {(x^2in ZZ), (x in ZZ), (1 in ZZ):}`

`-> x^2 + x + 1 in ZZ (dpcm)`.

Answer 2

\(A=\sqrt{\left(x^4+x^3+x^2\right)+\left(x^3+x^2+x\right)+\left(x^2+x+1\right)}\)

\(=\sqrt{x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+\left(x^2+x+1\right)}\)

\(=\sqrt{\left(x^2+x+1\right)^2}=\left|x^2+x+1\right|=x^2+x+1\)

\(\Rightarrow A\in Z\)