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b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{2}+2}-\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\) (vì \(\sqrt{5}\ge\sqrt{2}\)
=0
c) \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+\sqrt{3+1}\) (vì \(\sqrt{3}\ge1\))
\(=2\sqrt{3}\)
a)\(\sqrt{5+2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}-\sqrt{3-2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\) (vì \(\sqrt{3}\ge\sqrt{2}\))
=0
8) ĐKXĐ: $-2\leq x\leq 1$
PT $\Leftrightarrow (2x+4)-4\sqrt{2x+4}+4+[(1-x)-2\sqrt{1-x}+1]=0$
$\Leftrightarrow (\sqrt{2x+4}-2)^2+(\sqrt{1-x}-1)^2=0$
Dễ thấy: $(\sqrt{2x+4}-2)^2; (\sqrt{1-x}-1)^2\geq 0$ với mọi $x\in [-2;1]$ nên để tổng của chúng bằng $0$ thì:
$(\sqrt{2x+4}-2)^2=(\sqrt{1-x}-1)^2=0$
$\Leftrightarrow \sqrt{2x+4}=2; \sqrt{1-x}-1=0$
$\Leftrightarrow x=0$ (thỏa mãn)
Vậy.....
7)
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow x^2+[(x+1)-2\sqrt{x+1}+1]=0$
$\Leftrightarrow x^2+(\sqrt{x+1}-1)^2=0$
Ta thấy:
$x^2\geq 0; (\sqrt{x+1}-1)^2\geq 0$ với mọi $x\geq -1$
Do đó để tổng của chúng bằng $0$ thì $x^2=(\sqrt{x+1}-1)^2=0$
$\Leftrightarrow x=0$ (thỏa mãn)
Vậy.......
Câu a:
ĐKXĐ: \(x\neq \pm 3\)
\(\left|\frac{x+5}{-x^2+9}\right|=2\Rightarrow \left[\begin{matrix} \frac{x+5}{-x^2+9}=2\\ \frac{x+5}{-x^2+9}=-2\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x+5=2(-x^2+9)\\ x+5=-2(-x^2+9)\end{matrix}\right.\Rightarrow \left[\begin{matrix} 2x^2+x-13=0\\ 2x^2-x-23=0\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=\frac{-1\pm \sqrt{105}}{4}\\ x=\frac{1\pm \sqrt{185}}{4}\end{matrix}\right.\) (đều thỏa mãn )
Vậy.......
Câu b:
ĐKXĐ: \(x< 2\)
Ta có: \(\frac{4}{\sqrt{2-x}}-\sqrt{2-x}=2\)
\(\Rightarrow 4-(2-x)=2\sqrt{2-x}\)
\(\Leftrightarrow 4=(2-x)+2\sqrt{2-x}\)
\(\Leftrightarrow 5=(2-x)+2\sqrt{2-x}+1=(\sqrt{2-x}+1)^2\)
\(\Rightarrow \sqrt{2-x}+1=\sqrt{5}\) (do \(\sqrt{2-x}+1>0\) )
\(\Rightarrow \sqrt{2-x}=\sqrt{5}-1\)
\(\Rightarrow 2-x=6-2\sqrt{5}\)
\(\Rightarrow x=-4+2\sqrt{5}\) (thỏa mãn)
Vậy...........
\(A=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}-\dfrac{1}{\sqrt{a}-2}\)
=\(\dfrac{\left(\sqrt{a}+2\right).\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{\left(\sqrt{a}-4\right).\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
Điều kiện bạn tự ghi nhé 
\(B=\dfrac{1}{\sqrt{a}+1}:\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}+\dfrac{\sqrt{a}+2}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}+1}:\left(\dfrac{\left(\sqrt{a}+3\right).\left(\sqrt{a}-3\right)-\left(\sqrt{a}-2\right).\left(\sqrt{a}+2\right)+\sqrt{a}+2}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}+1}:\dfrac{a-9-a+4+\sqrt{a}+2}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{1}{\sqrt{a}+1}:\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-3\right).\left(\sqrt{a}-2\right)}\)
\(=\dfrac{1}{\sqrt{a}+1}:\dfrac{1}{\sqrt{a}-2}\)
\(=\dfrac{1}{\sqrt{a}+1}.\dfrac{\sqrt{a}-2}{1}=\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\)
\(\dfrac{\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\dfrac{\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\dfrac{\left|\sqrt{3}+\sqrt{2}\right|+\left|\sqrt{5}-\sqrt{3}\right|}{\left|\sqrt{5}+\sqrt{2}\right|}=\dfrac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}=1\)