Question

[Thị Thiệm Lê]

Giải phương trình \(\sqrt{1-2x}+\sqrt{1+2x}=2-x^2\)

Answer 1

[Nguyễn Việt Lâm]

ĐKXĐ: \(-\dfrac{1}{2}\le x\le\dfrac{1}{2}\)

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

\(\Leftrightarrow2+2\sqrt{1-4x^2}=\left(2-x^2\right)^2\)

Đặt \(\sqrt{1-4x^2}=t\ge0\Rightarrow x^2=\dfrac{1-t^2}{4}\)

Pt trở thành:

\(2+2t=\left(2-\dfrac{1-t^2}{4}\right)^2\)

\(\Leftrightarrow\left(t^2+7\right)^2=32\left(t+1\right)\)

\(\Leftrightarrow t^4+14t^2-32t+17=0\)

\(\Leftrightarrow\left(t-1\right)^2\left(t^2+2t+17\right)=0\)

\(\Leftrightarrow t=1\Rightarrow\sqrt{1-4x^2}=1\Rightarrow x=0\)