\(A=\left|2-\sqrt{7}\right|+7-2\sqrt{7}+1\)

\(=\sqrt{7}-2+8-2\sqrt{7}\) \(=6-\sqrt{7}\)

\(B=3\cdot1,5-4\cdot\left|3-\sqrt{2}\right|\) \(=4,5-4\left(3-\sqrt{2}\right)\)

\(=4,5-12+4\sqrt{2}\) \(=4\sqrt{2}-7,5\) 

Ta có: \(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)

\(=\sqrt{7}-2+8-2\sqrt{7}\)

\(=6-\sqrt{7}\)

\(\left(4+\sqrt{7}\right)\cdot\dfrac{\sqrt{4-\sqrt{7}}}{\sqrt{4+\sqrt{7}}}\)

\(=\left(4+\sqrt{7}\right)\cdot\dfrac{\sqrt{7}-1}{\sqrt{7}+1}\)

\(=\dfrac{\left(\sqrt{7}+1\right)^2\cdot\left(\sqrt{7}-1\right)}{\sqrt{7}+1}\cdot\dfrac{1}{2}\)

\(=\dfrac{6}{2}=3\)

\(=\dfrac{\left(8+2\sqrt{7}\right)\sqrt{8-2\sqrt{7}}}{2\sqrt{8+2\sqrt{7}}}=\dfrac{\left(\sqrt{7}+1\right)^2\sqrt{\left(\sqrt{7}-1\right)^2}}{2\sqrt{\left(\sqrt{7}+1\right)^2}}\)

\(=\dfrac{\left(\sqrt{7}+1\right)^2\left(\sqrt{7}-1\right)}{2\left(\sqrt{7}+1\right)}=\dfrac{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}{2}\)

\(=\dfrac{7-1}{2}=3\)

a) \(\sqrt{2}\left(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\right)\)

\(=\sqrt{2\cdot\left(4+\sqrt{7}\right)}+\sqrt{2\cdot\left(4-\sqrt{7}\right)}\)

\(=\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}\right)^2+2\cdot\sqrt{7}\cdot1+1^2}+\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot1+1^2}\)

\(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)

\(=\left|\sqrt{7}+1\right|+\left|\sqrt{7}-1\right|\)

\(=\sqrt{7}+1+\sqrt{7}-1\)

\(=2\sqrt{7}\)

b) \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

\(=\dfrac{\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2\cdot\left(2-\sqrt{3}\right)}-\sqrt{2\cdot\left(2+\sqrt{3}\right)}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}}{\sqrt{2}}\)

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

\(=\dfrac{\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{ }\)

\(=-\dfrac{2}{\sqrt{2}}\)

\(=-\sqrt{2}\)

a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)

\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)

\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)

b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)

\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)

\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)

b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)

c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)

\(D=a^{\dfrac{7}{2}}.a^{\dfrac{1}{3}}.a^{\dfrac{7}{4}}=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}=\sqrt[12]{a^{67}}\)

\(D=a^{\sqrt{2}-1}.a^{2\sqrt{2}}.a^{3-3\sqrt{2}}=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{3}}=a^2\)

\(D=\left(\sqrt{a}\right)^7\cdot\left(\sqrt[3]{a}\right)\left(\sqrt[4]{a}\right)^7\)

\(=a^{\dfrac{1}{2}\cdot7}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}\cdot7}\)

\(=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}\)

b: \(D=a^{\sqrt{2}-1}\cdot\left(a^2\right)^{\sqrt{2}}\cdot\left(a^3\right)^{1-\sqrt{2}}\)

\(=a^{\sqrt{2}-1}\cdot a^{2\sqrt{2}}\cdot a^{3-3\sqrt{2}}\)

\(=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{2}}=a^2\)

a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}=4\)

b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

\(=2+\sqrt{3}-2+\sqrt{3}\)

\(=2\sqrt{3}\)

c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)

\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)

\(A=\frac{\sqrt{7-2\sqrt{10}}.\left(7+2\sqrt{10}\right)\left(74-22\sqrt{10}\right)}{\sqrt{125}-4\sqrt{50}+5\sqrt{20}+\sqrt{8}}\)

\(=\frac{\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}.\left(78-6\sqrt{10}\right)}{5\sqrt{5}-20\sqrt{2}+10\sqrt{5}+2\sqrt{2}}\)

\(=\frac{\left(\sqrt{5}-\sqrt{2}\right)\left(78-6\sqrt{10}\right)}{15\sqrt{5}-18\sqrt{2}}\)

\(=\frac{\left(\sqrt{5}-\sqrt{2}\right)\left(26-2\sqrt{10}\right)}{5\sqrt{5}-6\sqrt{2}}\)

\(=\frac{30\sqrt{5}-36\sqrt{2}}{5\sqrt{5}-6\sqrt{2}}=6\)

1)\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{26^2}=\sqrt{5}-2+26=24-\sqrt{5}\)

2) \(=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)

3) \(=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)\(=\left[{}\begin{matrix}1\left(x>1\right)\\-1\left(x< 1\right)\end{matrix}\right.\)

4) \(=\sqrt{\left(\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}=2\sqrt{\dfrac{1}{2}}=\sqrt{2}\)

2. \(\dfrac{x^2-5}{x+\sqrt{5}}=\dfrac{x^2-\left(\sqrt{5}\right)^2}{x+\sqrt{5}}=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)

3. \(\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{x^2-2.x.1+1^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{|x-1|}{x-1}=\left[{}\begin{matrix}x-1>0\left(x>1\right)\\x-1< 0\left(x< 1\right)\end{matrix}\right.=\left[{}\begin{matrix}=1\\=\dfrac{x+1}{x-1}\end{matrix}\right.\)