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\(\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}.x\right)\)
\(\frac{x}{2}-\frac{3x-13}{5}=-\frac{14+7x}{10}\)
\(\frac{5x}{10}-\frac{2\left(3x-13\right)}{10}=-\frac{14+7x}{10}\)
\(\frac{5x-6x-26}{10}=\frac{-14-7x}{10}\)
\(\Rightarrow10.\left(5x-6x-26\right)=\left(-14-7x\right).10\)
\(\Rightarrow50x-60x-260=-140-70x\)
\(\Rightarrow-10x-260=-140-70x\)
\(\Rightarrow-10x+70x=-140+260\)
\(\Rightarrow60x=120\)
\(\Rightarrow x=2\)
\(\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
\(\Leftrightarrow\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
\(\Leftrightarrow\frac{x}{2}-\frac{3x}{5}=-\frac{7}{10}x-4\)
\(\Leftrightarrow-\frac{7}{10}x-\frac{x}{2}+\frac{3x}{5}=4\)
\(\Leftrightarrow x\left(-\frac{7}{10}-\frac{1}{2}+\frac{3}{5}\right)=4\)
\(\Leftrightarrow x\left(-\frac{3}{5}\right)=4\)
\(\Leftrightarrow x=-\frac{20}{3}\)
\(\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=\dfrac{7}{5}+\dfrac{7}{10}x\\ \Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}+\dfrac{13}{5}=\dfrac{7}{5}+\dfrac{7x}{10}\\ \Rightarrow\dfrac{5x}{10}-\dfrac{6x}{10}+\dfrac{26}{10}=\dfrac{14}{10}+\dfrac{7x}{10}\\ \Rightarrow\dfrac{5x}{10}-\dfrac{6x}{10}-\dfrac{7x}{10}=\dfrac{14}{10}-\dfrac{26}{10}\\ \Rightarrow\dfrac{-8x}{10}=-\dfrac{12}{10}\\ \Rightarrow\dfrac{-4x}{5}=-\dfrac{6}{5}\\ \Rightarrow-20x=-30\\ \Rightarrow x=\dfrac{3}{2}\)
a: 2x-3/2+3/4=-4
=>2x-3/4=-4
=>2x=-13/4
hay x=-13/8
b: \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\cdot\left(\dfrac{-3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)
\(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{-29}{6}=\dfrac{-2}{5}\cdot\dfrac{6}{29}=\dfrac{-12}{145}\)
=>2/3x+3/5=12/145
=>2/3x=-15/29
hay x=-45/58
c: \(\dfrac{x}{2}-\left(\dfrac{3}{5}x-\dfrac{13}{5}\right)=-\left(\dfrac{7}{10}x+\dfrac{7}{5}\right)\)
=>1/2x-3/5x+13/5=-7/10x-7/5
=>-1/10x+7/10x=-7/5-13/5
=>3/5x=-2
hay x=-2:3/5=-10/3
=>1/2x-3/5x+13/5=7/5+7/10x
=>-1/10x-7/10x=7/5-13/5
=>-4/5x=-6/5
=>x=6/5:4/5=6/5x5/4=6/4=1,5
a)\(-\frac{3}{2}-2x+\frac{3}{4}=2\)
\(\Leftrightarrow-2x=2+\frac{3}{2}-\frac{3}{4}=\frac{11}{4}\)
\(\Leftrightarrow x=\frac{11}{4}:\left(-2\right)=-\frac{11}{8}\)
b , pt <=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\) <=> \(\frac{3}{5}x\) = -4 <=> x = \(\frac{-20}{3}\)
c , \(\frac{2}{3x}-\frac{3}{12}\) = \(\frac{4}{5}-\left(\frac{7}{x}-x\right)\) (ĐK : x khác 0 )
<=> \(\frac{23}{3x}\) - x =\(\frac{21}{20}\) <=>\(-3x^2-\frac{21}{20}.3x+23=0\Leftrightarrow\left[{}\begin{matrix}x=\\x=\end{matrix}\right.\)...
`x/2-(3x/5-13/5)=7/5-7/10x`
`=>x/2-3x/5+13/5=7/5+(7x)/10`
`=>(5x)/10-(6x)/10-(7x)/10=7/5-13/5`
`=>(5x-6x-7x)/10=(7-13)/5`
`=>(-4x)/5=-6/5`
`=>4x=6`
`=>x=3/2`
Bài 1 :
a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)
b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)
\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)
\(=-1+\left(-2\right)=-1-2=-3\)
c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
Bài 2 :
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)
=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)
b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)
=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)
=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)
c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>
\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
a)Ta có: \(\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}x\right)\)
\(\Leftrightarrow\dfrac{x}{2}-\dfrac{3x-13}{5}=\dfrac{-7}{5}-\dfrac{7x}{10}\)
\(\Leftrightarrow\dfrac{5x}{10}-\dfrac{2\left(3x-13\right)}{10}=\dfrac{-14}{10}-\dfrac{7x}{10}\)
\(\Leftrightarrow5x-6x+26=-14-7x\)
\(\Leftrightarrow-x+26+14+7x=0\)
\(\Leftrightarrow6x=-40\)
hay \(x=-\dfrac{20}{3}\)
d) Ta có: \(\dfrac{2x-3}{2}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{6}+\dfrac{-9}{6}=\dfrac{5-3x}{6}-\dfrac{2}{6}\)
\(\Leftrightarrow6x-9-9=5-3x-2\)
\(\Leftrightarrow6x-18-3+3x=0\)
\(\Leftrightarrow x=\dfrac{7}{3}\)