Đáp án:
e) \(\left[ \begin{array}{l}
x = 8\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)5\left( {2x - 1} \right) + 4\left( {8 - x} \right) = - 5\\
\to 10x - 5 + 32 - 4x = - 5\\
\to 6x = - 32\\
\to x = - \dfrac{{16}}{3}\\
b)4x(x - 1) - 3({x^2} - 5) - {x^2} = x - 3 - (x + 4)\\
\to 4{x^2} - 4x - 3{x^2} + 15 - {x^2} = x - 3 - x - 4\\
\to - 4x = - 22\\
\to x = \dfrac{{11}}{2}\\
c)25{x^2} - 9 = 0\\
\to \left( {5x - 3} \right)\left( {5x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
5x - 3 = 0\\
5x + 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{5}\\
x = - \dfrac{3}{5}
\end{array} \right.\\
d)2(x + 3) - {x^2} - 3x = 0\\
\to 2\left( {x + 3} \right) - x\left( {x + 3} \right) = 0\\
\to \left( {x + 3} \right)\left( {2 - x} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\\
e){x^2} - 10x + 16 = 0\\
\to {x^2} - 8x - 2x + 16 = 0\\
\to x\left( {x - 8} \right) - 2\left( {x - 8} \right) = 0\\
\to \left( {x - 8} \right)\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 8\\
x = 2
\end{array} \right.
\end{array}\)
