tìm x, biết:
a, 8x ²-4x=0
b, 5x(x-3)+7(x-3)
c, 7x(x-2004)- x+2004=0
d, 2x ²=x
e, x ²-5x=0
f, x ²+5x=0
g, (x+1) ² - (x+1)(x-1)=0
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2 câu trả lời 981
\[\begin{array}{l}
a,{\rm{ }}8{x^2} - 4x = 0\\
\Leftrightarrow 4x\left( {2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
4x = 0\\
2x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \frac{1}{2}
\end{array} \right.\\
vay:S = \{ 0;\frac{1}{2}\} \\
b,{\rm{ }}5x\left( {x - 3} \right) + 7\left( {x - 3} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {5x + 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
5x + 7 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = \frac{{ - 7}}{5}
\end{array} \right.\\
vay:S = \{ 3;\frac{{ - 7}}{5}\} \\
c,{\rm{ }}7x\left( {x - 2004} \right) - {\rm{ }}x + 2004 = 0\\
\Leftrightarrow 7x\left( {x - 2004} \right) - \left( {x - 2004} \right) = 0\\
\Leftrightarrow \left( {x - 2004} \right)\left( {7x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2004 = 0\\
7x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2004\\
x = \frac{1}{7}
\end{array} \right.\\
Vay:S = \{ 2004;\frac{1}{7}\} \\
d,{\rm{ }}2{x^2} = x\\
\Leftrightarrow 2{x^2} - x = 0\\
\Leftrightarrow x\left( {2x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
2x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \frac{1}{2}
\end{array} \right.\\
vay:S = \{ 0;\frac{1}{2}\} \\
e,{\rm{ }}x{{\rm{ }}^2} - 5x = 0\\
\Leftrightarrow x\left( {x - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 5
\end{array} \right.\\
vay:S = \{ 0;5\} \\
f,{\rm{ }}{x^2} + 5x = 0\\
\Leftrightarrow x\left( {x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 5
\end{array} \right.\\
Vay:S = \{ 0; - 5\} \\
g,{\rm{ }}{\left( {x + 1} \right)^2}{\rm{ }} - {\rm{ }}\left( {x + 1} \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {x + 1 - x + 1} \right) = 0\\
\Leftrightarrow 2\left( {x + 1} \right) = 0\\
\Leftrightarrow x + 1 = 0\\
\Leftrightarrow x = - 1\\
vay:x = - 1
\end{array}\]
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