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<=>\(\left|\dfrac{1}{2}-2x\right|=\dfrac{5}{3}< =>\left[{}\begin{matrix}\dfrac{1}{2}-2x=\dfrac{5}{3}\\\dfrac{1}{2}-2x=\dfrac{-5}{3}\end{matrix}\right.< =>\left[{}\begin{matrix}2x=\dfrac{-7}{6}\\2x=\dfrac{13}{6}\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{-7}{12}\\x=\dfrac{13}{12}\end{matrix}\right.\)
\(\left|\dfrac{1}{2}+2x\right|+\dfrac{2}{3}=\dfrac{7}{3}\)
\(\left|\dfrac{1}{2}+2x\right|=\dfrac{7}{3}-\dfrac{2}{3}=\dfrac{5}{3}\)
⇔\(\left[{}\begin{matrix}\dfrac{1}{2}+2x=\dfrac{5}{3}\\\dfrac{1}{2}+2x=-\dfrac{5}{3}\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{7}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
Vậy ...
a)
\(6x+5=4x-7\)
\(\Leftrightarrow6x-4x=-7-5\)
\(\Leftrightarrow2x=-12\)
\(\Leftrightarrow x=-6\)
b)
\(-3\left(x-5\right)-1=2x-1\)
\(\Leftrightarrow-3x+15-1=2x-1\)
\(\Leftrightarrow-3x-2x=-1-15+1\)
\(\Leftrightarrow-5x=-15\)
\(\Leftrightarrow x=3\)
q, bạn ghi đề rõ nhé
s, \(\dfrac{2}{3}x=\dfrac{1}{2}-\dfrac{7}{12}=\dfrac{-1}{12}\Leftrightarrow x=-\dfrac{1}{12}:\dfrac{2}{3}=-\dfrac{3}{24}=-\dfrac{1}{8}\)
t, \(\dfrac{3}{4}x=\dfrac{1}{6}-\dfrac{1}{5}=\dfrac{-1}{30}\Leftrightarrow x=-\dfrac{1}{30}:\dfrac{3}{4}=-\dfrac{4}{90}=-\dfrac{2}{45}\)
u, \(\dfrac{1}{6}x=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{1}{8}:\dfrac{1}{6}=\dfrac{6}{8}=\dfrac{3}{4}\)
s, 23x=12−712=−112⇔x=−112:23=−324=−1823x=12−712=−112⇔x=−112:23=−324=−18
t, 34x=16−15=−130⇔x=−130:34=−490=−24534x=16−15=−130⇔x=−130:34=−490=−245
u, 16x=38−14=18⇔x=18:16=68=34
HT
Ta có : \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\Rightarrow\frac{3}{2x+1}+\frac{5.2}{2\left(2x+1\right)}-\frac{3.2}{3\left(2x+1\right)}=\frac{6}{13}\)
=> \(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
=> \(\frac{3+5-2}{2x+1}=\frac{6}{13}\)
=> \(\frac{6}{2x+1}=\frac{6}{13}\)
=> 2x + 1 = 13
=> 2x = 12
=> x = 6
Vậy x = 6
\(\frac{3}{2x+1}+\frac{10}{2\left(2x+1\right)}-\frac{6}{3\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
\(\Rightarrow2x+1=13\left(6=6\right)\)
\(2x=12\)
\(x=6\)
\(\left(x-3\right)\cdot\left(y-5\right)=3\)
=>\(\left(x-3\right)\cdot\left(y-5\right)=1\cdot3=3\cdot1=\left(-1\right)\cdot\left(-3\right)=\left(-3\right)\cdot\left(-1\right)\)
=>\(\left(x-3;y-5\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(4;8\right);\left(6;6\right);\left(2;2\right);\left(0;4\right)\right\}\)
\(\frac{3}{4}x-\frac{1}{7}=\frac{1}{6}x+\frac{1}{3}\)
=> \(\frac{3}{4}x-\frac{1}{7}-\frac{1}{6}x-\frac{1}{3}=0\)
=> \(\left(\frac{3}{4}-\frac{1}{6}\right)x+\left(-\frac{1}{7}-\frac{1}{3}\right)=0\)
=> \(\frac{7}{12}x-\frac{10}{21}=0\)
=> \(\frac{7}{12}x=\frac{10}{21}\)
=> x = 40/49
\(\frac{3}{4}x-\frac{1}{7}=\frac{1}{6}x+\frac{1}{3}\)
\(\Rightarrow\frac{3}{4}x-\frac{1}{6}x=\frac{1}{3}+\frac{1}{7}\)
\(\Rightarrow\left(\frac{3}{4}-\frac{1}{6}\right)x=\frac{10}{21}\)
\(\Rightarrow\frac{7}{12}x=\frac{10}{21}\)
\(\Rightarrow x=\frac{10}{21}:\frac{7}{12}\)
\(\Rightarrow x=\frac{40}{49}\)