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a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
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A/ \(\frac{4}{5}+\frac{1}{5}x=\left(-\frac{1}{28}\right)\)
\(\frac{1}{5}x=\left(-\frac{1}{28}\right)-\frac{4}{5}\)
\(\frac{1}{5}x=\left(-\frac{5}{140}\right)-\frac{112}{140}\)
\(\frac{1}{5}x=-\frac{117}{140}\)
\(x=\left(-\frac{117}{140}\right):\frac{1}{5}\)
\(x=\left(-\frac{117}{28}\right)\)
B/ \(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\)
\(\frac{3x}{7}+1=\left(-\frac{1}{28}\right).\left(-4\right)\)
\(\frac{3x}{7}+1=\frac{1}{7}\)
\(\frac{3x}{7}=\frac{1}{7}-1\)
\(\frac{3x}{7}=\frac{1}{7}-\frac{7}{7}\)
\(\frac{3x}{7}=-\frac{6}{7}\)
\(\Rightarrow3x=\left(-6\right)\)(theo cách l8)
\(\Rightarrow x=\left(-6\right):3=\left(-2\right)\)
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a. \(\frac{4}{5}+\frac{1}{5}x=\frac{1}{28}\)
\(\Leftrightarrow\frac{1}{5}x=\frac{1}{28}-\frac{4}{5}\)
\(\Leftrightarrow\frac{1}{5}x=\frac{5}{140}-\frac{112}{140}\)
\(\Leftrightarrow\frac{1}{5}x=\frac{-107}{140}\)
\(\Leftrightarrow x=\frac{-107}{140}\div\frac{1}{5}\)
\(\Leftrightarrow x=\frac{-107}{140}\times5\)
\(\Leftrightarrow x=\frac{-107}{28}=\)
b. \(\left(\frac{3x}{7}+1\right)\div\left(-4\right)=\frac{-1}{28}\)
\(\Leftrightarrow\left(\frac{3x}{7}+\frac{7}{7}\right)=\frac{-1}{28}\times\left(-4\right)\)
\(\Leftrightarrow\frac{3x+1}{7}=\frac{1}{7}\)
\(\Leftrightarrow3x+1=1\)
\(\Leftrightarrow3x=1-1\)
\(\Leftrightarrow3x=0\Leftrightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\frac{3x}{7+1}:\left(-4\right)=\frac{-1}{28}\)
\(\frac{3x}{7+1}\)=\(\frac{-1}{28}.\left(-4\right)\)
\(\frac{3x}{7+1}\)=\(\frac{1}{7}\)
\(\frac{3x}{8}\)=\(\frac{1}{7}\)
\(3x\)=\(\frac{8.1}{7}\)=\(\frac{8}{7}\)
\(x\)=\(\frac{8}{7}\): \(3\)
\(x\)=\(\frac{8}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(4\left(x-2\right)-2\left(x+3\right)=-28\)
\(\Leftrightarrow4x-8-2x-6+28=0\)
\(\Leftrightarrow2x+14=0\)
\(\Leftrightarrow2x=-14\)
hay x=-7
Vậy: x=-7
b) Ta có: \(3x+7-9x=-11\)
\(\Leftrightarrow-6x+7+11=0\)
\(\Leftrightarrow-6x+18=0\)
\(\Leftrightarrow-6x=-18\)
hay x=3
Vậy: x=3