1) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\) ĐKXĐ \(x\ge1\)

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

ta có:

\(\left|\sqrt{x-1}+1\right|+\left|1-\sqrt{x-1}\right|\)\(\ge\left|\sqrt{x-1}+1+1-\sqrt{x-1}\right|=2\)

dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}+1\ge0\\1-\sqrt{x-1}\ge0\end{matrix}\right.\Leftrightarrow x\le2\)

kết hợp vs điều kiện ta có nghiệm của phương trình \(\left\{x|1\le x\le2\right\}\)

câu 2,4 tương tự

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

đề rút gọn \(A=x+2y-\sqrt{x^2-4xy+4y^2}\) biết \(x\ge2y\)

\(A=x+2y-\sqrt{x^2-4xy+4y^2}\)

\(A=x+2y-\sqrt{\left(x-2y\right)^2}\)

\(A=x+2y-x+2y=4y\) (do \(x\ge2y\))

a, \(A=\dfrac{\sqrt{x+\sqrt{2x-1}}}{\sqrt{2}}\) DKXD: \(x\ge\dfrac{1}{2}\)

\(A=\dfrac{\sqrt{2x+2\sqrt{2x-1}}}{2}=\dfrac{\sqrt{2x-1}+1}{2}\)( vi \(x\ge\dfrac{1}{2}\))

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

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

\(B=-2x+2\)

a, \(\sqrt{9x+9}-4\sqrt{\dfrac{x+1}{4}}=5\) \(x\ge-1\)

\(\Leftrightarrow3\sqrt{x+1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow x+1=25\Leftrightarrow x=24\)