Đáp án:

\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = \sqrt {\sqrt 3  + \sqrt 2 } \\
x =  - \sqrt {\sqrt 3  + \sqrt 2 } 
\end{array} \right.\\
g)\left[ \begin{array}{l}
x = \sqrt {\sqrt 3 } \\
x =  - \sqrt {\sqrt 3 } 
\end{array} \right.\\
e)\left[ \begin{array}{l}
x = \sqrt 5 \\
x =  - \sqrt 5 
\end{array} \right.\\
h)x \in \emptyset \\
f)\left[ \begin{array}{l}
x = \sqrt {2 - \sqrt 5 } \\
x =  - \sqrt {2 - \sqrt 5 } 
\end{array} \right.\\
i)\left[ \begin{array}{l}
x = \dfrac{9}{4}\\
x =  - \dfrac{1}{4}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a){x^2} = \sqrt 3  + \sqrt 2 \\
 \to \left[ \begin{array}{l}
x = \sqrt {\sqrt 3  + \sqrt 2 } \\
x =  - \sqrt {\sqrt 3  + \sqrt 2 } 
\end{array} \right.\\
g){x^2} = \sqrt 3 \\
 \to \left[ \begin{array}{l}
x = \sqrt {\sqrt 3 } \\
x =  - \sqrt {\sqrt 3 } 
\end{array} \right.\\
e){x^2} = 5\\
 \to \left[ \begin{array}{l}
x = \sqrt 5 \\
x =  - \sqrt 5 
\end{array} \right.\\
h)2{x^2} = 2\sqrt 3  - 3\sqrt 2 \\
 \to {x^2} = \dfrac{{2\sqrt 3  - 3\sqrt 2 }}{2}\\
Do:\dfrac{{2\sqrt 3  - 3\sqrt 2 }}{2} < 0\\
 \to x \in \emptyset \\
f){x^2} = 2 - \sqrt 5 \\
 \to \left[ \begin{array}{l}
x = \sqrt {2 - \sqrt 5 } \\
x =  - \sqrt {2 - \sqrt 5 } 
\end{array} \right.\\
i){\left( {x - 1} \right)^2} = \dfrac{{25}}{{16}}\\
 \to \left| {x - 1} \right| = \dfrac{5}{4}\\
 \to \left[ \begin{array}{l}
x - 1 = \dfrac{5}{4}\\
x - 1 =  - \dfrac{5}{4}
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{9}{4}\\
x =  - \dfrac{1}{4}
\end{array} \right.
\end{array}\)

Đáp án:

`d)x^2-sqrt3=sqrt2`

`<=>x^2=sqrt2+sqrt3`

`<=>` \(\left[ \begin{array}{l}x=\sqrt{\sqrt2+\sqrt3}\\x=-\sqrt{\sqrt2+\sqrt3}\end{array} \right.\) 

`e)x^2-5=0`

`<=>x^2=5`

`<=>` \(\left[ \begin{array}{l}x=\sqrt5\\x=-\sqrt5\end{array} \right.\) 

`f)x^2+sqrt5=2`

`<=>x^2=2-sqrt5`

Vì `2=sqrt4<sqrt5`

`=>2-sqrt5<0`

Mà `x^2>=0`

Vậy pt vô nghiệm

`g)x^2=sqrt3`

`<=>` \(\left[ \begin{array}{l}x=\sqrt[4]{3}\\x=-\sqrt[4]{3}\end{array} \right.\) 

`h)2x^2+3sqrt2=2sqrt3`

`<=>2x^2=2sqrt3-3sqrt2`

Vì `2sqrt3=sqrt{12}<sqrt{18}=3sqrt2`

`=>2sqrt3-3sqrt2<0`

Mà `x^2>=0`

Vậy pt vô nghiệm

`i)(x-1)^2=1 9/16`

`<=>(x-1)^2=25/16`

`<=>` \(\left[ \begin{array}{l}x-1=\dfrac54\\x-1=-\dfrac54\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac94\\x=-\dfrac14\end{array} \right.\)