Đáp án:

\(\begin{array}{l}
B5:\\
A = 1\\
B = 1\\
C =  - 1
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
B5:\\
A = \sqrt {\left( {8 + 3\sqrt 7 } \right)\left( {8 - 3\sqrt 7 } \right)} \\
 = \sqrt {64 - {{\left( {3\sqrt 7 } \right)}^2}} \\
 = \sqrt {64 - 63}  = 1\\
B = \sqrt {2 + \sqrt 3 } .\sqrt {\left( {2 + \sqrt {2 + \sqrt 3 } } \right)\left( {2 - \sqrt {2 + \sqrt 3 } } \right)} \\
 = \sqrt {2 + \sqrt 3 } .\sqrt {\left( {4 - {{\left( {\sqrt {2 + \sqrt 3 } } \right)}^2}} \right)} \\
 = \sqrt {2 + \sqrt 3 } .\sqrt {4 - 2 - \sqrt 3 } \\
 = \sqrt {2 + \sqrt 3 } .\sqrt {2 - \sqrt 3 } \\
 = \sqrt {\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)} \\
 = \sqrt {4 - 3}  = 1\\
C = \left( {1 + \sqrt 2 } \right).\left( {{{\sqrt 2 }^2} - {{\left( {\sqrt {1 + \sqrt 2 } } \right)}^2}} \right)\\
 = \left( {1 + \sqrt 2 } \right).\left( {2 - 1 - \sqrt 2 } \right)\\
 = \left( {1 + \sqrt 2 } \right).\left( {1 - \sqrt 2 } \right)\\
 = 1 - 2 =  - 1
\end{array}\)

Bài 4: Rút gọn các biểu thức:

a.

$\frac{\sqrt[]{3+\sqrt[]{5}}}{\sqrt[]{2}}$

=$\frac{\sqrt[]{(3+\sqrt[]{5}).2}}{2}$

=$\frac{\sqrt[]{6+2\sqrt[]{5}}}{2}$

=$\frac{\sqrt[]{(\sqrt[]{5}+1)²}}{2}$

=$\frac{|\sqrt[]{5}+1|}{2}$

=$\frac{\sqrt[]{5}+1}{2}$

b.

$\frac{\sqrt[]{x-\sqrt[]{2x-1}}}{\sqrt[]{2}}$ (với x>1)
=$\frac{\sqrt[]{(x-\sqrt[]{2x-1}).2}}{2}$
=$\frac{\sqrt[]{2x-2\sqrt[]{2x-1}}}{2}$ 
=$\frac{\sqrt[]{(\sqrt[]{2x-1}-1)²}}{2}$
=$\frac{|\sqrt[]{2x-1}-1|}{2}$
=$\frac{\sqrt[]{2x-1}-1}{2}$

c.

$\frac{\sqrt[]{3x+\sqrt[]{6x-1}}}{\sqrt[]{2}}$ (x>$\frac{1}{6}$)
=$\frac{\sqrt[]{6x+2\sqrt[]{6x-1}}}{2}$
=$\frac{\sqrt[]{(\sqrt[]{6x-1}+1)²}}{2}$
= $\frac{|\sqrt[]{6x-1}+1|}{2}$ 
=$\frac{\sqrt[]{6x-1}+1}{2}$ 

Bài 5: Tính:

A=$\sqrt[]{8+3\sqrt[]{7}}$.$\sqrt[]{8-3\sqrt[]{7}}$
 =$\sqrt[]{(8+3\sqrt[]{7}).(8-3\sqrt[]{7})}$ 
 =$\sqrt[]{8²-(3\sqrt[]{7})²}$ 
 =$\sqrt[]{64-63}$ 
 =$\sqrt[]{1}$ 
 =1

B=$\sqrt[]{2+\sqrt[]{3}}$.$\sqrt[]{2+\sqrt[]{2+\sqrt[]{3}}}$.$\sqrt[]{2-\sqrt[]{2+\sqrt[]{3}}}$ 
 =$\sqrt[]{2+\sqrt[]{3}}$.$\sqrt[]{2²-(\sqrt[]{2+\sqrt[]{3}})²}$
 =$\sqrt[]{2+\sqrt[]{3}}$.$\sqrt[]{4-2-\sqrt[]{3}}$ 
 =$\sqrt[]{2+\sqrt[]{3}}$.$\sqrt[]{2-\sqrt[]{3}}$ 
 =$\sqrt[]{2²-(\sqrt[]{3})²}$ 
 =1

C=(1+$\sqrt[]{2}$).($\sqrt[]{2}$+$\sqrt[]{1+\sqrt[]{2}}$).($\sqrt[]{2}$-$\sqrt[]{1+\sqrt[]{2}}$)
 =(1+$\sqrt[]{2}$).[($\sqrt[]{2}$)²-($\sqrt[]{1+\sqrt[]{2}}$)²]
 =(1+$\sqrt[]{2}$).(2-1-$\sqrt[]{2}$)
 =(1+$\sqrt[]{2}$).(1-$\sqrt[]{2}$)
 =1²-($\sqrt[]{2}$)²
 =-1

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