Question

Bài 1 : Nếu : 

\(\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)=1\) \(=1\)

Tìm x

Answer 1

(-8+x^2)^5=1

<=>-8+x^2=1

<=>x^2=9

<=>x=3 hoặc -3

Vậy x=3 hoặc -3

Answer 2

Ta có: \(\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)=1\)

    \(\Leftrightarrow\left(-8+x^2\right)^5=1\)

    \(\Leftrightarrow x^2-8=\pm1\)

 + \(x^2-8=1\)\(\Leftrightarrow\)\(x^2=9\)\(\Leftrightarrow\)\(x=\pm3\)

 + \(x^2-8=-1\)\(\Leftrightarrow\)\(x^2=7\)\(\Leftrightarrow\)\(x=\pm\sqrt{7}\) 

Vậy \(S=\left\{-3,-\sqrt{7},\sqrt{7},3\right\}\)