a) \(ĐKXĐ:x\ge0\)

\(x=\sqrt{x}\)\(\Leftrightarrow x-\sqrt{x}=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)( thỏa mãn )

Vậy \(x=0\)hoặc \(x=1\)

b) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)

\(\frac{4}{x-1}+\frac{2}{x}=\frac{3x+4}{x^2-x}\)\(\Leftrightarrow\frac{4x}{x\left(x-1\right)}+\frac{2\left(x-1\right)}{x\left(x-1\right)}=\frac{3x+4}{x\left(x-1\right)}\)

\(\Rightarrow4x+2\left(x-1\right)=3x+4\)

\(\Leftrightarrow4x+2x-2=3x+4\)

\(\Leftrightarrow4x+2x-3x=4+2\)

\(\Leftrightarrow3x=6\)\(\Leftrightarrow x=2\)( thỏa mãn )

Vậy \(x=2\)

Sai đề mọi người ưiii

Sửa đê: đổi \(\sqrt{2}\)thành \(2\)

\(ĐKXĐ:x\ge1\)

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

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

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

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

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

TH1: Nếu \(\sqrt{x-1}-1\le0\)\(\Leftrightarrow\sqrt{x-1}\le1\)\(\Leftrightarrow x-1\le1\)\(\Leftrightarrow x\le2\)( phải thỏa mãn cả ĐKXĐ )

\(\Rightarrow\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\)

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

\(\Leftrightarrow2=2\)( luôn đúng )

TH2: Nếu \(\sqrt{x-1}-1\ge0\)\(\Leftrightarrow\sqrt{x-1}\ge1\)\(\Leftrightarrow x-1\ge1\)\(\Leftrightarrow x\ge2\)

\(\Rightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)

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

\(\Leftrightarrow2\sqrt{x-1}=2\)\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=2\)( thỏa mãn )

Vậy \(x=2\)

\(3x+2-\sqrt{9x^2+6x+1}\)

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

\(=3x+2-\left(3x+1\right)\)(Chú ý \(x>\frac{1}{3}\))

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

Giúp mình với ạ

cách của symbolab:

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

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

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

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

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

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

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

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

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

\(-x^2+2=\frac{x^2}{1-4x+4x^2}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1+\sqrt{3}}{2}\end{cases}}\)