Question

Giải phương trình:
a,\(\sqrt{x^2-4}-\sqrt{x-2}=0\)
b,\(\sqrt{x^2-9}-\sqrt{x+3}=0\)

Answer 1

a) ĐK: \(x\ge2\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(chon\right)\\x=-1\left(loai\right)\end{matrix}\right.\)

b) ĐK: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

\(pt\Leftrightarrow\sqrt{x^2-9}=\sqrt{x+3}\)

\(\Leftrightarrow x^2-9=x+3\)

\(\Leftrightarrow x^2-x-12=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(chon\right)\\x=-3\left(chon\right)\end{matrix}\right.\)

Vậy...