Question

Giải phương trình

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

 

Answer 1

TH1: `x>=0`

`\sqrt(x^2-2x+1)=x^2-x`

`<=> |x-1| =x^2-x`

TH1.1: `x-1 >=0 <=> x>=1`

`x-1=x^2-x`

`<=>x=1` (TM)

TH1.2: `x>= 0 \vee x<1`

`=>` VN.

TH2: `x<0`

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

`<=> |x+1|=x^2+x`

TH2.1: `x>=-1`

`x+1=x^2+x`

`<=> x=\pm 1`

TH2.2: `x<-1`

`-x-1=x^2+x`

`<=>x=-1`

Vậy `S={-1;1}`.

Answer 2

Ta có \(x^2-\left|x\right|\ge0\Leftrightarrow\left|x\right|^2\ge\left|x\right|\Leftrightarrow\left|x\right|\ge1\).

Khi đó \(PT\Leftrightarrow\sqrt{\left(\left|x\right|-1\right)^2}=x^2-\left|x\right|\)

\(\Leftrightarrow\left|x\right|-1=x^2-\left|x\right|\Leftrightarrow\left(\left|x\right|-1\right)^2=0\Leftrightarrow\left|x\right|=1\Leftrightarrow x=\pm1\) (thỏa mãn |x| \(\geq \) 1).

Vậy...