Đáp án:

 10) \(\sqrt {x + 1}  + 1\)

Giải thích các bước giải:

\(\begin{array}{l}
5)E = \dfrac{{\sqrt {4 - 2\sqrt 3 } }}{{\sqrt 2 }} = \dfrac{{\sqrt {3 - 2\sqrt 3 .1 + 1} }}{{\sqrt 2 }}\\
 = \dfrac{{\sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}} }}{{\sqrt 2 }} = \dfrac{{\sqrt 3  - 1}}{{\sqrt 2 }}\\
6)F = \dfrac{{\sqrt {8 - 2\sqrt {15} } }}{{\sqrt 2 }} = \dfrac{{\sqrt {5 - 2.\sqrt 5 .\sqrt 3  + 3} }}{{\sqrt 2 }}\\
 = \dfrac{{\sqrt {{{\left( {\sqrt 5  - \sqrt 3 } \right)}^2}} }}{{\sqrt 2 }} = \dfrac{{\sqrt 5  - \sqrt 3 }}{{\sqrt 2 }}\\
7)G = \sqrt {\dfrac{{8 - 2\sqrt {15} }}{4}}  + \sqrt {\dfrac{{4 - 2\sqrt 3 }}{2}}  - \dfrac{{3\sqrt 5  - 3}}{2} + \sqrt 5 \\
 = \dfrac{{\sqrt {{{\left( {\sqrt 5  - \sqrt 3 } \right)}^2}} }}{2} + \dfrac{{\sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}} }}{2} - \dfrac{{3\sqrt 5  - 3}}{2} + \sqrt 5 \\
 = \dfrac{{\sqrt 5  - \sqrt 3  + \sqrt 3  - 1 - 3\sqrt 5  + 3 + 2\sqrt 5 }}{2}\\
 = \dfrac{2}{2} = 1\\
8)H = {\left( {\dfrac{{\sqrt {4 + 2\sqrt 3 } }}{{\sqrt 2 }} + \dfrac{{\sqrt {6 - 2\sqrt 5 } }}{{\sqrt 2 }} - \dfrac{{\sqrt 5 }}{{\sqrt 2 }}} \right)^2}\\
 = {\left( {\dfrac{{\sqrt {3 + 2\sqrt 3 .1 + 1}  + \sqrt {5 - 2.\sqrt 5 .1 + 1}  - \sqrt 5 }}{{\sqrt 2 }}} \right)^2}\\
 = {\left( {\dfrac{{\sqrt {{{\left( {\sqrt 3  + 1} \right)}^2}}  + \sqrt {{{\left( {\sqrt 5  - 1} \right)}^2}}  - \sqrt 5 }}{{\sqrt 2 }}} \right)^2}\\
 = {\left( {\dfrac{{\sqrt 3  + 1 + \sqrt 5  - 1 - \sqrt 5 }}{{\sqrt 2 }}} \right)^2}\\
 = {\left( {\dfrac{{\sqrt 3 }}{{\sqrt 2 }}} \right)^2} = \dfrac{3}{2}\\
9)G = \sqrt {x - 1 + 2\sqrt {x - 1} .1 + 1} \\
 = \sqrt {{{\left( {\sqrt {x - 1}  + 1} \right)}^2}} \\
 = \left| {\sqrt {x - 1}  + 1} \right|\\
10)H = \sqrt {x + 1 + 2\sqrt {x + 1} .1 + 1} \\
 = \sqrt {{{\left( {\sqrt {x + 1}  + 1} \right)}^2}} \\
 = \sqrt {x + 1}  + 1
\end{array}\)

Đáp án:

Giải thích các bước giải:

5.$E=$$\sqrt[]{2-\sqrt[]{3}}$$=$$\frac{\sqrt[]{4-2\sqrt[]{3}}}{\sqrt[]{2}}$$=$$\frac{\sqrt[]{(\sqrt[]{3}-1)^2}}{\sqrt[]{2}}$$=$$\frac{\sqrt[]{6}-\sqrt[]{2}}{2}$

6.$F$=$\sqrt[]{4-\sqrt[]{15}}$ =$\sqrt[]{\frac{8-2\sqrt[]{15}}{2}}$ = $\frac{\sqrt[]{3}+\sqrt[]{5}}{\sqrt[]{2}}$

7.$G$=$\sqrt[]{\frac{4-\sqrt[]{15}}{2}}$ + $\sqrt[]{\frac{2-\sqrt[]{3}}{2}}$ - $\frac{\sqrt[]{45}-3}{2}$ +$\sqrt[]{5}$ =$\frac{\sqrt[]{5}-\sqrt[]{3}}{2}$+$\frac{\sqrt[]{3}-1}{2}$-$\frac{\sqrt[]{45}-3}{2}$ +$\sqrt[]{5}$=1-$\sqrt[]{5}$ +$\sqrt[]{5}$ =1

8.h=$($$\sqrt[]{2+\sqrt[]{3}}$$ + $$\sqrt[]{3-\sqrt[]{5}}$$ -$$\sqrt[]{\frac{5}{2}}$$)^{2}$ =$($$\frac{\sqrt[]{3}+1+\sqrt[]{5}-1}{\sqrt[]{2}}$-$\sqrt[]{\frac{5}{2}}$ )$^2$=($\frac{\sqrt[]{3}}{\sqrt[]{2}}$)$^2$=$\frac{3}{2}$

9. SORRy mik ko bt làm bài này nha kho quá

10.H=$\sqrt[]{x+2+2\sqrt[]{x+1}}$ =$\sqrt[]{(\sqrt[]{x+1}+1)^2}$=$\sqrt[]{x+1}+1$