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\(x=\dfrac{4}{27}-\dfrac{2}{3}\)
\(x=-\dfrac{14}{27}\)
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\(x^3=\frac{27}{8}=\left(\frac{3}{2}\right)^3=>x=\frac{3}{2}\)
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a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
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Vì ( 2x - 4).( 3x + 9 ) = 0
=> 2x - 4 hoặc 3x + 9 = 0
TH1: 2x - 4 = 0
2x = 4
x=2
TH2: 3x + 9 = 0
3x = -9
x = -3
Vậy, x = -3 hoặc 2
\(\left(2x-4\right)\left(3x+9\right)=0\)
\(\orbr{\begin{cases}2x-4=0\\3x+9=0\end{cases}}\)
TH1. \(2x-4=0\)
\(2x=4\)
\(x=2\)
TH2. \(3x+9=0\)
\(3x=9\)
\(x=3\)
Vậy \(x\in\left\{2;3\right\}\)
@Nghệ Mạt
#cua
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\(\frac{x-2}{3}=\frac{27}{x-2}ĐKx\ne2\)
\(\Leftrightarrow\left(x-2\right)^2=81\Leftrightarrow x-2=\pm9\)
TH1 : \(x-2=9\Leftrightarrow x=11\)
TH2 : \(x-2=-9\Leftrightarrow x=-7\)