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Trả lời :
a, \(\frac{3}{4}-\left(\frac{1}{2}\div x+\frac{1}{2}\right)=\frac{3}{5}\)
=> \(\frac{1}{2}\div x+\frac{1}{2}=\frac{3}{20}\)
=> \(\frac{1}{2}\div x=\frac{-7}{20}\)
=> \(x=\frac{-10}{7}\)
b, (4 - x) . (2x + 3) = 0
=> \(\orbr{\begin{cases}4-x=0\\2x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=\frac{-3}{2}\end{cases}}\)
c, \(\frac{4}{-3}=\frac{-12}{x}\)
=> 4x = 36
=> x = 9
d, \(\frac{4x}{-3}=\frac{12}{-x}\)
=> \(-4x^2=-36\)
=> 4x2 = 36
=> x2 = 9
=> x = \(\pm3\)
a) \(\frac{3}{4}-\left(\frac{1}{2}:x+\frac{1}{2}\right)=\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{3}{4}-\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{15}{20}-\frac{12}{20}\)
\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{13}{20}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{13}{20}-\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{13}{20}-\frac{10}{20}\)
\(\Leftrightarrow\frac{1}{2}:x=\frac{3}{20}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{3}{20}\)
\(\Leftrightarrow x=\frac{1}{2}.\frac{20}{3}=\frac{10}{3}\)
Vậy: \(x=\frac{10}{3}\)
b) \(3x.\left(\frac{1}{2}.x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\\frac{1}{2}x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
Vậy: \(x\in\left\{0;2\right\}\)
c) \(\left(4-x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\2x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=\frac{3}{2}\end{cases}}}\)
Vậy: \(x\in\left\{4;\frac{3}{2}\right\}\)
d) \(\frac{4}{-3}=\frac{-12}{x}\)
\(\Leftrightarrow4x=\left(-12\right).\left(-3\right)\)
\(\Leftrightarrow4x=36\)
\(\Leftrightarrow x=9\)
Vậy: \(x=9\)
e) \(\frac{4x}{-3}=\frac{12}{-x}\)
\(\Leftrightarrow4x.\left(-x\right)=12.\left(-3\right)\)
\(\Leftrightarrow-4x^2=-36\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy: \(x\in\left\{3;-3\right\}\)
a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)
\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)
=>2x+10=0
hay x=-5
b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)
=>x-2017=0
hay x=2017