Lời giải:
a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+5\sqrt{x}=-20$

$\Leftrightarrow 5\sqrt{x}-8\sqrt{2x}=-20$

$\Leftrightarrow \sqrt{x}(5-8\sqrt{2})=-20$

$\Leftrightarrow \sqrt{x}=\frac{20}{8\sqrt{2}-5}$

$\Rightarrow x=(\frac{20}{8\sqrt{2}-5})^2$

b. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 3\sqrt{5x}-5\sqrt{3x}+4\sqrt{x}=10$

$\Leftrightarrow \sqrt{x}(3\sqrt{5}-5\sqrt{3}+4)=10$

$\Leftrightarrow \sqrt{x}=\frac{10}{3\sqrt{5}-5\sqrt{3}+4}$

$\Rightarrow x=(\frac{10}{3\sqrt{5}-5\sqrt{3}+4})^2$

\(a,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\\ =\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\=2\sqrt{2} \)

\(b,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)

\(c,=x-4+\sqrt{\left(4^2-2.4.x+x^2\right)}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4=2x-8\)    (vì \(x>4\) )

@seven 

\(a,ĐK:x\ge-7\\ PT\Leftrightarrow\sqrt{\left(\sqrt{x+7}+1\right)^2}+\sqrt{x+7-\sqrt{x+7}-6}=4\)

Đạt \(\sqrt{x+7}=a\ge0\)

\(PT\Leftrightarrow\sqrt{\left(a+1\right)^2}+\sqrt{a^2-a-6}=4\\ \Leftrightarrow a+1+\sqrt{a^2-a-6}=4\\ \Leftrightarrow\sqrt{a^2-a-6}=3-a\\ \Leftrightarrow a^2-a-6=a^2-6a+9\\ \Leftrightarrow5a=15\Leftrightarrow a=3\\ \Leftrightarrow\sqrt{x+7}=3\\ \Leftrightarrow x+7=9\\ \Leftrightarrow x=2\left(tm\right)\)

\(2.< =>5\sqrt{x-1}-6\sqrt{x-1}-3\sqrt{x-1}=2\sqrt{2x-3}\)

\(< =>\sqrt{x-1}\left(5-6+3\right)=2\sqrt{2x-3}\)

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

\(< =>x-1=2x-3\)

\(< =>x=2\)