Question

 

giúp mình với ạ

Answer 1

\(\Leftrightarrow x^4-2x^2+x^2-2=0\)

\(\Leftrightarrow x^2-2=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

Answer 2

\(x^4-x^2-2=0\Leftrightarrow\left(x^2-2\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x^2-2=0\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\)(do \(x^2+1>0\forall x\))

Answer 3

<=>\(\left(x^2-2\right)\left(x^2+1\right)=0\)

<=>\(x^2-2=0\)(do \(x^2+1\)>0 với mọi x)

<=>\(x^2=2\)

<=>x=\(\pm\sqrt{2}\)

Answer 4

\(\Leftrightarrow x^4-2x^2+x^2-2=0\)

\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(x^2+1\right)=0\)

Ta có: \(x^2+1\ge0\) với mọi x nên

\(x^2-2=0\Leftrightarrow x=\sqrt{2}\)

Answer 5

Đặt x^2 = t ( t >= 0 ) 

t^2 - t - 2 = 0 

,=> t^2 - 2t + t - 2 = 0 

<=> t(t-2) + (t-2) = 0 <=> (t+1)(t-2) = 0 <=> t = -1 (ktm) ; t = 2 

=> x= \(\pm\sqrt{2}\)

 

Answer 6

\(x^4-x^2-2=0\\ \Leftrightarrow x^4+x^2-\left(2x^2-2\right)=0\\ \Leftrightarrow x^2.\left(x^2+1\right)-2.\left(x^2+1\right)=0\\ \Leftrightarrow\left(x^2-2\right).\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+1=0\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=\pm\sqrt{2}\\ \Rightarrow S=\left\{\pm2\right\}\)