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a) \(3\dfrac{4}{5}:\dfrac{8}{5}=0,25:x\)
\(\Rightarrow\dfrac{19}{5}.\dfrac{5}{8}=\dfrac{x}{4}\)
\(\Rightarrow\dfrac{x}{4}=2\Rightarrow x=8\)
b) \(2x+\dfrac{3}{24}=3x-\dfrac{1}{32}\)
\(\Rightarrow x=\dfrac{1}{8}+\dfrac{1}{32}=\dfrac{5}{32}\)
c) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)
\(\Rightarrow221x-34=152x+380\)
\(\Rightarrow69x=414\Rightarrow x=6\)
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a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)
hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)
b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)
nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)
hay \(x=\dfrac{8}{41}\)
c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|2x-1\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
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a)\(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(4x^2-20x-\left(4x^2-7x+3\right)=5\)
\(4x^2-20x-4x^2+7x-3=5\)
\(-13x=8\)
\(x=-\frac{8}{13}\)
b)\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(48x^2-32x+5+3x-48x^2-7+112x=81\)
\(83x-2=81\)
\(x=1\)
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a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
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\(2x-3+3|x-1|=4x+1.\)
\(\Leftrightarrow3|x-1|=2x+4\)
*Với x < 1 ta có phương trình:
\(3\left(-x+1\right)=2x+4\)
\(\Leftrightarrow-3x+3=2x+4\)
\(\Leftrightarrow5x+1=0\)
\(\Leftrightarrow x=-\frac{1}{5}\)(TM)
*Với \(x\ge1\)ta có phương trình:
\(2x-3+3\left(x-1\right)=4x+1\)
\(\Leftrightarrow2x-3+3x-3=4x+1\)
\(\Leftrightarrow x-7=0\)
\(\Leftrightarrow x=7\)(TM)
Vậy ............