Question

Tìm x, biết:

a) \(\sqrt{x^2+1}-x=3\)

b) \(\sqrt{1-x>x}\)

Answer 1

a) \(\sqrt{x^2+1}+x=3\)

Nhận xét \(x^2+1\ge0\forall x\)

\(\Leftrightarrow\left|x^2+1\right|+x=3\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\left(TM\right)\)

Vậy \(S=\left\{-2;1\right\}\)

b) \(\sqrt{1-x}>x\)

ĐK \(x\le1\)

\(\Leftrightarrow\left|1-x\right|>x\)

\(\Leftrightarrow1-x>x\Leftrightarrow1>2x\)

\(\Rightarrow x< \dfrac{1}{2}\)

Vậy \(x< \dfrac{1}{2}\) thì \(\sqrt{1-x}>x\)