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

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

Tất nhiên là \(=0\)

\(a,\sqrt{25x^2}=10\)

\(\sqrt{\left(5x\right)^2}=10\)

\(5x=10\)

\(x=2\)

b. <=> \(\sqrt{4\left(x^2-1\right)}=2\sqrt{15}\)     ĐKXĐ: x>=1,x>=-1

<=> \(4\left(x^2-1\right)=60\Leftrightarrow x^2-1=15\Leftrightarrow x^2-16=0\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\)

<=>x=+-4

\(A=\left(\sqrt{5}-\sqrt{2}\right)^2+2\sqrt{10}=\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^2+2\sqrt{10}\)

=\(5-2\sqrt{10}+2+2\sqrt{10}=7\)

câu b hình như sai đề

\(C=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)=\sqrt{3-\sqrt{5}}\sqrt{2}\left(\sqrt{5}-1\right)\)

\(=\sqrt{2\left(3-\sqrt{5}\right)}\left(\sqrt{5}-1\right)=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\)

=\(\sqrt{5-2\sqrt{5}+1}\left(\sqrt{5}-1\right)=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\)

=\(\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)=\left(\sqrt{5}-1\right)^2=\left(\sqrt{5}\right)^2-2\sqrt{5}+1\)

=\(6-2\sqrt{5}\)

\(A=\left(\sqrt{5}-\sqrt{2}\right)^2+2\sqrt{10}=\left(\sqrt{5.1}-\sqrt{2.1}\right)^2+2\sqrt{10}=7-2\sqrt{10}+2\sqrt{10}=7\)        

Câu B hình như có cái gì đó không ổn          

\(C=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)=\sqrt{3-\sqrt{5}}\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2\left(3-\sqrt{5}\right)}\left(\sqrt{5-1}\right)=\sqrt{6-2\sqrt{5}}\left(-1+\sqrt{5}\right)=6-2\sqrt{5}\)