mong mọi người giải giúp em vs 

c: ĐKXĐ: \(\begin{cases}x-2\ge0\\ x+2\ge0\\ x^2-4\ge0\end{cases}\)

=>x>=2 và \(x^2\ge4\)

=>x>=2

Ta có: \(\sqrt{x-2}-\sqrt{x+2}=2\cdot\sqrt{x^2-4}-2x+2\)

=>\(\sqrt{x-2}-\sqrt{x+2}+2=2\cdot\sqrt{x^2-4}-2x+4\)

=>\(\sqrt{x-2}-\frac{x+2-4}{\sqrt{x+2}+2}=2\cdot\sqrt{\left(x-2\right)\left(x+2\right)}-2\left(x-2\right)\)

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

=>\(\sqrt{x-2}\left(1-\frac{\sqrt{x-2}}{\sqrt{x+2}+2}-2\sqrt{x+2}+4\right)=0\)

=>\(\sqrt{x-2}=0\)

=>x-2=0

=>x=2(nhận)

a: ĐKXĐ: \(x^2-1\ge0\)

=>x>=1 hoặc x<=-1

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

=>\(\sqrt{x-\sqrt{x^2-1}}-1+\sqrt{x+\sqrt{x^2-1}}-1=0\)

=>\(\frac{x-\sqrt{x^2-1}-1}{\sqrt{x-\sqrt{x^2-1}+1}}+\frac{x+\sqrt{x^2-1}-1}{\sqrt{x+\sqrt{x^2-1}+1}}=0\)

=>\(\frac{\sqrt{x-1}\left(\sqrt{x-1}-\sqrt{x+1}\right)}{\sqrt{x-\sqrt{x^2-1}+1}}+\frac{\sqrt{x-1}\left(\sqrt{x+1}+\sqrt{x-1}\right)}{\sqrt{x+\sqrt{x^2-1}+1}}=0\)

=>\(\sqrt{x-1}\left(\frac{\left(\sqrt{x-1}-\sqrt{x+1}\right)}{\sqrt{x-\sqrt{x^2-1}+1}}+\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)}{\sqrt{x+\sqrt{x^2-1}+1}}\right)=0\)

=>\(\sqrt{x-1}=0\)

=>x-1=0

=>x=1(nhận)

a, ĐK: \(x\ge1\)

Đặt \(\sqrt{5x-1}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)

\(pt\Leftrightarrow\left(a+b\right)\left(\dfrac{a^2+b^2}{2}-ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2=2\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a-b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=b+2\end{matrix}\right.\)

TH1: \(a=b\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}\Leftrightarrow x=0\left(l\right)\)

TH2: \(a=b+2\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}+2\)

\(\Leftrightarrow5x-1=x-1+4+4\sqrt{x-1}\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}=0\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}+1=1\)

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

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

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

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