a: \(\left(1+\sqrt3-\sqrt2\right)\left(1+\sqrt3-\sqrt2\right)\)

\(=\left(1+\sqrt3-\sqrt2\right)^2\)

\(=1+3+2+2\sqrt3-2\sqrt2-2\sqrt6=6+2\sqrt3-2\sqrt2-2\sqrt6\)

b: \(\left(1+\frac{5-\sqrt5}{1-\sqrt5}\right)\left(\frac{5+\sqrt5}{\sqrt5+1}+1\right)\)

\(=\left(1-\frac{\sqrt5\left(\sqrt5-1\right)}{\sqrt5-1}\right)\left(\frac{\sqrt5\left(\sqrt5+1\right)}{\sqrt5+1}+1\right)\)

\(=\left(1-\sqrt5\right)\left(1+\sqrt5\right)=1-5=-4\)

c: \(\frac{1}{\sqrt3-\sqrt2}\cdot\frac{1}{\sqrt3-\sqrt2}=\frac{1}{\left(\sqrt3-\sqrt2\right)^2}\)

\(=\frac{1}{5-2\sqrt6}=\frac{5+2\sqrt6}{\left(5-2\sqrt6\right)\left(5+2\sqrt6\right)}=\frac{5+2\sqrt6}{25-24}=5+2\sqrt6\)

\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)

1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)

2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)

3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)

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

5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)

6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)

7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)

\( \left( {\sqrt 2 + \sqrt 3 + \sqrt 5 } \right)\left( {\sqrt 2 + \sqrt 3 - \sqrt 5 } \right)\left( {\sqrt 2 - \sqrt 3 + \sqrt 5 } \right)\left( { - \sqrt 2 + \sqrt 3 + \sqrt 5 } \right)\\ = \left[ {{{\left( {\sqrt 2 + \sqrt 3 } \right)}^2} - 5} \right]\left( {\sqrt 5 + \sqrt 2 - \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 2 + \sqrt 3 } \right)\\ = \left( {2 + 2\sqrt 6 + 3 - 5} \right)\left[ {5 - {{\left( {\sqrt 2 - \sqrt 3 } \right)}^2}} \right]\\ = 2\sqrt 6 \left[ {5 - \left( {2 - 2\sqrt 6 + 3} \right)} \right]\\ = 2\sqrt 6 .2\sqrt 6 \\ = 4\sqrt {36} = 4.6 = 24 \)

24

e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

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

=2

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

\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)

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

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

\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}\)

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

g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)

=-(căn 5+2)(căn 5-2)

=-(5-4)=-1

h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)

=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6

=4

\(c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)

\(=\sqrt{4+5\sqrt{3}+5\left(5-\sqrt{3}\right)}\)

\(=\sqrt{4+5\sqrt{3}+25-5\sqrt{3}}\)

\(=\sqrt{29}\)

a: \(A=\sqrt{27}-2\sqrt{48}+3\sqrt{75}\)

\(=3\sqrt3-2\cdot4\sqrt3+3\cdot5\sqrt3\)

\(=3\sqrt3-8\sqrt3+15\sqrt3=18\sqrt3-8\sqrt3=10\sqrt3\)

b: \(B=\sqrt{\left(\sqrt5-2\right)^2}-\sqrt{\left(\sqrt5-3\right)^2}\)

\(=\sqrt5-2-\left|3-\sqrt5\right|\)

\(=\sqrt5-2-3+\sqrt5=-5+2\sqrt5\)

c: \(C=\sqrt{\left(2\sqrt3+1\right)^2}+\sqrt{\left(2\sqrt3-5\right)^2}\)

\(=2\sqrt3+1+\left|5-2\sqrt3\right|\)

\(=2\sqrt3+1+5-2\sqrt3\)

=6

d: \(D=\sqrt{9-4\sqrt5}-\sqrt{14+6\sqrt5}\)

\(=\sqrt{\left(\sqrt5-2\right)^2}-\sqrt{\left(3+\sqrt5\right)^2}\)

\(=\sqrt5-2-\left(3+\sqrt5\right)=\sqrt5-2-3-\sqrt5=-5\)

e: \(E=\frac{4}{\sqrt5-2}-\frac{32}{\sqrt5+1}\)

\(=\frac{4\left(\sqrt5+2\right)}{\left(\sqrt5-2\right)\left(\sqrt5+2\right)}-\frac{32\left(\sqrt5-1\right)}{\left(\sqrt5+1\right)\left(\sqrt5-1\right)}\)

\(=4\left(\sqrt5+2\right)-8\left(\sqrt5-1\right)=4\sqrt5+8-8\sqrt5+8=-4\sqrt5+16\)

f: \(M=\frac{10}{3\sqrt2-4}+\frac{28}{3\sqrt2+2}\)

\(=\frac{10\left(3\sqrt2+4\right)}{\left(3\sqrt2+4\right)\left(3\sqrt2-4\right)}+\frac{28\left(3\sqrt2-2\right)}{\left(3\sqrt2-2\right)\left(3\sqrt2+2\right)}\)

\(=\frac{10\left(3\sqrt2+4\right)}{2}+\frac{28\left(3\sqrt2-2\right)}{18-4}=5\left(3\sqrt2+4\right)+2\left(3\sqrt2-2\right)=15\sqrt2+20+6\sqrt2-4=21\sqrt2+16\)