Question

giảipt

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

Answer 1

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\(\Leftrightarrow\sqrt{1-x}+\sqrt{1+x}+2+2\sqrt{1-x^2}-6=0\)

Đặt \(\sqrt{1-x}+\sqrt{1+x}=a>0\Leftrightarrow a^2=2+2\sqrt{1-x^2}\)

\(\Rightarrow a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)

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

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

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

Answer 2

Nguyễn Việt Lâm helppp mkk

Answer 3

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

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

Đặt \(t=\sqrt{1-x}+\sqrt{1+x}\left(t\ge0\right)\)

\(\Rightarrow t^2=2+2\sqrt{1-x^2}\Rightarrow2\sqrt{1-x^2}=t^2-2\)

\(Pt\Leftrightarrow\) \(t^2+t-6=0\Leftrightarrow\) \(\left[{}\begin{matrix}t=2\\t=-3\left(L\right)\end{matrix}\right.\)

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

\(\Leftrightarrow2+2\sqrt{1-x^2}=2\Leftrightarrow\)\(x^2=1\Leftrightarrow x=\pm1\)