Question

Giải phương trình: \(8-x^2=4\left(\sqrt{1-x}+\sqrt{1+x}\right)\)

Answer 1

ĐK: \(-1\le x\le1\)

Đặt \(t=\sqrt{1-x}+\sqrt{1+x}\left(\sqrt{2}\le t\le2\right)\)

\(pt\Leftrightarrow7+\dfrac{t^4-4t^2+4}{4}=4t\)

\(\Leftrightarrow t^4-4t^2-16t+32=0\)

\(\Leftrightarrow\left(t-2\right)\left(t^3+2t-16\right)=0\)

\(\Leftrightarrow t=2\) (Vì \(t\le2\Rightarrow t^3+2t-16\le-4\))

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

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

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

\(\Leftrightarrow x=0\left(tm\right)\)