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a: \(\Leftrightarrow\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{5}{3}=\dfrac{1}{6}\\x\cdot\dfrac{5}{3}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}:\dfrac{5}{3}=\dfrac{3}{30}=\dfrac{1}{10}\\x=-\dfrac{1}{10}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x-1\right|=\dfrac{3}{2}:\dfrac{3}{4}=2\)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
c: \(\Leftrightarrow\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)
\(\Leftrightarrow x+\dfrac{3}{5}=\dfrac{7}{3}-x\)
=>2x=44/15
hay x=22/15
\(\frac{x-3}{x+3}=\frac{3}{7}\Leftrightarrow\left(x-3\right).7=3.\left(x+3\right)\)
\(\Leftrightarrow7x-21=3x+9\Rightarrow7x-3x=9-\left(-21\right)\)
\(\Leftrightarrow4x=30;x=7,5\)
b) \(\frac{x+4}{20}=\frac{5}{x+4}\Leftrightarrow\left(x+4\right)\left(x+4\right)=5.20\)
\(x^2+\left(4+4\right)^2=100\)
\(x^2=100-64=36\)
\(\Rightarrow x\in\left\{-6;6\right\}\)
a: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>\(\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
b: \(\Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
=>1/x-1=3/4
=>x-1=4/3
=>x=7/3
\(\left(x-3\right):\dfrac{4}{5}=20:\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right).\dfrac{5}{4}-\left(x-3\right).\dfrac{1}{20}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{5}{4}-\dfrac{1}{20}\right)=0\)
\(\Leftrightarrow\left(x-3\right).\dfrac{6}{5}=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)