Toán Lớp 8: Giải phương trình sau:
a, (1/2x-3) – 3/x(2x-3) = 5/x
b, (x+2/x-2) – 1/x = 2/x(x-2)
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Toán Lớp 8: Giải phương trình sau:
a, (1/2x-3) – 3/x(2x-3) = 5/x
b, (x+2/x-2) – 1/x = 2/x(x-2)
Comments ( 1 )
Giải đáp:
\(\begin{array}{l}
a)x = \dfrac{4}{3}\\
b)x = – 1
\end{array}\)
Lời giải và giải thích chi tiết:
\(\begin{array}{l}
a)DK:x \ne \left\{ {0;\dfrac{3}{2}} \right\}\\
\dfrac{1}{{2x – 3}} – \dfrac{3}{{x\left( {2x – 3} \right)}} = \dfrac{5}{x}\\
\to \dfrac{{x – 3 – 5\left( {2x – 3} \right)}}{{x\left( {2x – 3} \right)}} = 0\\
\to x – 3 – 10x + 15 = 0\\
\to – 9x + 12 = 0\\
\to x = \dfrac{4}{3}\\
b)DK:x \ne \left\{ {0;2} \right\}\\
\dfrac{{x + 2}}{{x – 2}} – \dfrac{1}{x} = \dfrac{2}{{x\left( {x – 2} \right)}}\\
\to \dfrac{{x\left( {x + 2} \right) – x + 2 – 2}}{{x\left( {x – 2} \right)}} = 0\\
\to {x^2} + 2x – x = 0\\
\to {x^2} + x = 0\\
\to x\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( l \right)\\
x = – 1
\end{array} \right.
\end{array}\)