Đáp án:

$\begin{array}{l}
1){\left( {2x - 1} \right)^3} = 8\\
 \Leftrightarrow {\left( {2x - 1} \right)^3} = {2^3}\\
 \Leftrightarrow 2x - 1 = 2\\
 \Leftrightarrow 2x = 3\\
 \Leftrightarrow x = \dfrac{3}{2}\\
Vậy\,x = \dfrac{3}{2}\\
{\left( {2x - 3} \right)^2} = 9\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 3\\
2x - 3 =  - 3
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 6\\
2x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 0
\end{array} \right.\\
Vậy\,x = 3;x = 0\\
2)\\
{x^2} = {x^5}\\
 \Leftrightarrow {x^5} - {x^2} = 0\\
 \Leftrightarrow {x^2}\left( {{x^3} - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
{x^3} = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
Vậy\,x = 0;x = 1\\
{\left( {3y - 1} \right)^{10}} = {\left( {3y - 1} \right)^{20}}\\
 \Leftrightarrow {\left( {3y - 1} \right)^{20}} - {\left( {3y - 1} \right)^{10}} = 0\\
 \Leftrightarrow {\left( {3y - 1} \right)^{10}}\left( {{{\left( {3y - 1} \right)}^{10}} - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
3y - 1 = 0\\
{\left( {3y - 1} \right)^{10}} = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
y = \dfrac{1}{3}\\
3y - 1 = 1\\
3y - 1 =  - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
y = \dfrac{1}{3}\\
y = \dfrac{2}{3}\\
y = 0
\end{array} \right.\\
Vậy\,y = 0;y = \dfrac{1}{3};y = \dfrac{2}{3}\\
{\left( {x - 5} \right)^2} = {\left( {1 - 3x} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 1 - 3x\\
x - 5 = 3x - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = 6\\
2x =  - 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x =  - 2
\end{array} \right.\\
Vậy\,x =  - 2;x = \dfrac{3}{2}
\end{array}$

Đáp án: + Giải thích các bước giải:

 `1) `

`a) (2x-1)^3=8`

`⇔(2x-1)^3=2^3`

`⇔2x-1=2`

`⇔2x=3`

`⇔x=3/2`

Vậy `x=3/2`

`b)(2x-3)^2=9`

`⇔`\(\left[ \begin{array}{l}2x-3=3\\2x-3=-3\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}2x=6\\2x=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=3\\x=0\end{array} \right.\) 

Vậy `x=3` hoặc `x=0`

`2)`

`a)x^2=x^5`

`⇔x^2-x^5=0`

`⇔x^2(1-x^3)=0`

`⇔`\(\left[ \begin{array}{l}x^2=0\\1-x^3=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\) 

Vậy `x=0` hoặc `x=1`

`b) (3y-1)^10=(3y-1)^20`

`⇔(3y-1)^10-(3y-1)^20=0`

`⇔(3y-1)^10 [1-(3y-1)^10]=0`

`⇔`\(\left[ \begin{array}{l}(3y-1)^10=0\\1-(3y-1)^10=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}3y-1=0\\(3y-1)^10=1\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}y=\dfrac{1}{3}\\3y-1=1\\3y-1=-1\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}y=\dfrac{1}{3}\\y=\dfrac{2}{3}\\y=0\end{array} \right.\) 

Vậy `y=1/3` hoặc `y=2/3` hoặc `y=0`

`c)(x-5)^2=(1-3x)^2`

`⇔`\(\left[ \begin{array}{l}x-5=1-3x\\x-5=-(1-3x)\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x-5=1-3x\\x-5=3x-1\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x+3x=5+1\\x-3x=5-1\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}4x=6\\-2x=4\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=\dfrac{3}{2}\\2x=-4\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=-2\end{array} \right.\) 

Vậy `x=3/2` hoặc `x=-2`