Đáp án:
l. \(\left[ \begin{array}{l}
x = - 9\\
x = \dfrac{1}{5}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.4x\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 3
\end{array} \right.\\
c.2x\left( {x - 17} \right) - \left( {x - 17} \right) = 0\\
\to \left( {x - 17} \right)\left( {2x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 17\\
x = \dfrac{1}{2}
\end{array} \right.\\
e.4{x^2} - 4x + 1 = 0\\
\to {\left( {2x - 1} \right)^2} = 0\\
\to 2x - 1 = 0\\
\to x = \dfrac{1}{2}\\
g.{x^2} = \dfrac{3}{{25}}\\
\to \left[ \begin{array}{l}
x = \dfrac{{\sqrt 3 }}{5}\\
x = - \dfrac{{\sqrt 3 }}{5}
\end{array} \right.\\
k.4{x^2} = {\left( {x + 4} \right)^2}\\
\to \left| {2x} \right| = \left| {x + 4} \right|\\
\to \left[ \begin{array}{l}
2x = x + 4\\
2x = - x - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
3x = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = - \dfrac{4}{3}
\end{array} \right.\\
b.7x\left( {x = 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.\\
d.6x\left( {x - 1999} \right) - \left( {x - 1999} \right) = 0\\
\to \left( {x - 1999} \right)\left( {6x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1999\\
x = \dfrac{1}{6}
\end{array} \right.\\
f.9 = 64{x^2}\\
\to \left| {8x} \right| = 3\\
\to \left[ \begin{array}{l}
8x = 3\\
8x = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{8}\\
x = - \dfrac{3}{8}
\end{array} \right.\\
h.7 = 16{x^2}\\
\to \left| {4x} \right| = \sqrt 7 \\
\to \left[ \begin{array}{l}
4x = \sqrt 7 \\
- 4x = \sqrt 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{\sqrt 7 }}{4}\\
x = - \dfrac{{\sqrt 7 }}{4}
\end{array} \right.\\
l.{\left( {3x + 4} \right)^2} = {\left( {2x - 5} \right)^2}\\
\to \left| {3x + 4} \right| = \left| {2x - 5} \right|\\
\to \left[ \begin{array}{l}
3x + 4 = 2x - 5\\
3x + 4 = - 2x + 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 9\\
5x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 9\\
x = \dfrac{1}{5}
\end{array} \right.
\end{array}\)
