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$\frac{x-1}{4}$ =$\frac{9}{x-1}$ $\frac{-1}{2}$ -$\frac{x-1}{3}$ =$\frac{3}{4}$ $\frac{1}{3}$-(x-$\frac{1}{2}$)^2= $\frac{-2}{3}$ 2x-$\frac{11}{3
Question
$\frac{x-1}{4}$ =$\frac{9}{x-1}$
$\frac{-1}{2}$ -$\frac{x-1}{3}$ =$\frac{3}{4}$
$\frac{1}{3}$-(x-$\frac{1}{2}$)^2= $\frac{-2}{3}$
2x-$\frac{11}{3}$ =$\frac{x}{2}$-$\frac{-5}{4}$
3.(x-$\frac{1}{2}$)+2.(x+$\frac{1}{4}$)=$\frac{-3}{2}$
$\frac{1}{2}$ .(x-3)-2.(x-$\frac{1}{3}$ )=$\frac{-5}{2}$
$\frac{1}{2}$ -2.(x-1)=$\frac{-1}{3}$
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Môn Toán
4 years
2021-04-28T03:19:57+00:00
2021-04-28T03:19:57+00:00 1 Answers
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Answers ( )
Đáp án:
c) \(\left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = – \dfrac{1}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne 1\\
\dfrac{{x – 1}}{4} = \dfrac{9}{{x – 1}}\\
\to {\left( {x – 1} \right)^2} = 36\\
\to \left| {x – 1} \right| = 6\\
\to \left[ \begin{array}{l}
x – 1 = 6\\
x – 1 = – 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 7\\
x = – 5
\end{array} \right.\\
b) – \dfrac{1}{2} – \dfrac{{x – 1}}{3} = \dfrac{3}{4}\\
\to \dfrac{{ – 6 – 4x + 4 – 9}}{{12}} = 0\\
\to – 4x – 11 = 0\\
\to x = – \dfrac{{11}}{4}\\
c)\dfrac{1}{3} – {\left( {x – \dfrac{1}{2}} \right)^2} = – \dfrac{2}{3}\\
\to 1 = {\left( {x – \dfrac{1}{2}} \right)^2}\\
\to \left| {x – \dfrac{1}{2}} \right| = 1\\
\to \left[ \begin{array}{l}
x – \dfrac{1}{2} = 1\\
x – \dfrac{1}{2} = – 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = – \dfrac{1}{2}
\end{array} \right.\\
d)2x – \dfrac{{11}}{3} = \dfrac{x}{2} + \dfrac{5}{4}\\
\to \dfrac{3}{2}x = \dfrac{{16}}{3}\\
\to x = \dfrac{{32}}{9}\\
e)3x – \dfrac{3}{2} + 2x + \dfrac{1}{2} = – \dfrac{3}{2}\\
\to 5x = – \dfrac{1}{2}\\
\to x = – \dfrac{1}{{10}}\\
f)\dfrac{1}{2}x – \dfrac{3}{2} – 2x + \dfrac{2}{3} = – \dfrac{5}{2}\\
\to – \dfrac{3}{2}x = – \dfrac{5}{3}\\
\to x = \dfrac{{10}}{9}\\
g)\dfrac{1}{2} – 2x + 2 = – \dfrac{1}{3}\\
\to – 2x = – \dfrac{{17}}{6}\\
\to x = \dfrac{{17}}{{12}}
\end{array}\)