Question

Giải các phương trình sau:

a) \(\sqrt{2-x}-\sqrt{x^2-4}=0\)

b) \(\sqrt{x^2-3x+1}=\sqrt{x+1}\)

Answer 1

\(a,\sqrt{2-x}-\sqrt{x^2-4}=0\)

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

\(\Leftrightarrow\left(\sqrt{2-x}\right)^2=\left(\sqrt{x^2-4}\right)^2\)

\(\Leftrightarrow2-x=x^2-4\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

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

\(\Leftrightarrow\left(\sqrt{x^2-3x+1}\right)^2=\left(\sqrt{x+1}\right)^2\)

\(\Leftrightarrow x^2-3x+1=x+1\)

\(\Leftrightarrow x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)