1, tìm x a, (3x-1)^2-25=0 b, 8x^3-50x=0 c,(x-2)(x^2+3x+7)+2(x^2-4)-5(x-2)=0 2, tìm x a, 3x(x-1)+x-1=0 b,2(x+3)-x^2-3x=0 c, 4x^2-25-(2x-5)(2x+7)=0 d, x^3+27+(x+3)(x-9)=0

Giải thích các bước giải:

$\begin{array}{l}
B1:\\
a,{(3x - 1)^2} - 25 = 0\\
 \Leftrightarrow {\left( {3x - 1} \right)^2} = 25\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - 1 = 5\\
3x - 1 =  - 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = 6\\
3x =  - 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \dfrac{{ - 4}}{3}
\end{array} \right.\\
b,8{x^3} - 50x = 0\\
 \Leftrightarrow 2x\left( {4{x^2} - 25} \right) = 0\\
 \Leftrightarrow 2x\left( {2x - 5} \right)\left( {2x + 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 0\\
2x - 5 = 0\\
2x + 5 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{5}{2}\\
x = \dfrac{{ - 5}}{2}
\end{array} \right.\\
c,(x - 2)({x^2} + 3x + 7) + 2({x^2} - 4) - 5(x - 2) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 3x + 7} \right) + 2\left( {x - 2} \right)\left( {x + 2} \right) - 5\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left[ {\left( {{x^2} + 3x + 7} \right) + 2\left( {x + 2} \right) - 5} \right] = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 5x + 6} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 2} \right)\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 2 = 0\\
x + 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 2\\
x =  - 3
\end{array} \right.\\
B2:\\
a,{\rm{ }}3x\left( {x - 1} \right) + x - 1 = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {3x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
3x + 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{ - 1}}{3}
\end{array} \right.\\
b,2(x + 3) - {x^2} - 3x = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {2 - x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2 - x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
x = 2
\end{array} \right.\\
c,4{x^2} - 25 - (2x - 5)(2x + 7) = 0\\
 \Leftrightarrow \left( {2x - 5} \right)\left( {2x + 5} \right) - \left( {2x - 5} \right)\left( {2x + 7} \right) = 0\\
 \Leftrightarrow \left( {2x - 5} \right)\left( {2x + 5 - \left( {2x + 7} \right)} \right) = 0\\
 \Leftrightarrow  - 2\left( {2x - 5} \right) = 0\\
 \Leftrightarrow 2x - 5 = 0\\
 \Leftrightarrow x = \dfrac{5}{2}\\
d,{x^3} + 27 + (x + 3)(x - 9) = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 3x + 9 + x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 2x} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right)x\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x = 0\\
x - 2 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
x = 0\\
x = 2
\end{array} \right.
\end{array}$