Question

a\(\sqrt{x-1}+2\sqrt{9x-9}-14=0\)

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

Answer 1

\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+6\sqrt{x-1}=14\\ \Leftrightarrow7\sqrt{x-1}=14\\ \Leftrightarrow\sqrt{x-1}=2\Leftrightarrow x-1=4\\ \Leftrightarrow x=5\left(tm\right)\\ b,ĐK:-2\le x\le2\\ PT\Leftrightarrow\sqrt{2-x}\left(1-\sqrt{2+x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2-x=0\\2+x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Answer 2

a) ĐKXĐ: \(x\ge1\)

\(pt\Leftrightarrow\sqrt{x-1}+6\sqrt{x-1}=14\)

\(\Leftrightarrow7\sqrt{x-1}=14\Leftrightarrow\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\)

b) ĐKXĐ: \(-2\le x\le2\)

\(pt\Leftrightarrow\sqrt{2-x}-\sqrt{\left(2-x\right)\left(x+2\right)}=0\)

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

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