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a: \(\Leftrightarrow\dfrac{7}{2}x-\dfrac{3}{4}=\dfrac{1}{2}x+\dfrac{5}{2}\)
\(\Leftrightarrow3x=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{10}{4}+\dfrac{3}{4}=\dfrac{13}{4}\)
=>x=13/12
b: \(\Leftrightarrow x\cdot\left(\dfrac{2}{3}-\dfrac{1}{2}\right)=-\dfrac{1}{3}+\dfrac{2}{5}\)
\(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{-5+6}{15}=\dfrac{1}{15}\)
\(\Leftrightarrow x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)
c: \(\Leftrightarrow x\cdot\dfrac{1}{3}+x\cdot\dfrac{2}{5}+\dfrac{2}{5}=0\)
\(\Leftrightarrow x\cdot\dfrac{11}{15}=-\dfrac{2}{5}\)
\(\Leftrightarrow x=-\dfrac{2}{5}:\dfrac{11}{15}=\dfrac{-2}{5}\cdot\dfrac{15}{11}=\dfrac{-30}{55}=\dfrac{-6}{11}\)
d: \(\Leftrightarrow-\dfrac{1}{3}x+\dfrac{1}{2}+\dfrac{2}{3}-x-\dfrac{1}{2}=5\)
\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{2}{3}=5\)
\(\Leftrightarrow-\dfrac{4}{3}x=5-\dfrac{2}{3}=\dfrac{13}{3}\)
\(\Leftrightarrow x=\dfrac{13}{3}:\dfrac{-4}{3}=\dfrac{-13}{4}\)
e: \(\Leftrightarrow\left(\dfrac{x+2015}{5}+1\right)+\left(\dfrac{x+2016}{4}+1\right)=\left(\dfrac{x+2017}{3}+1\right)+\left(\dfrac{x+2018}{2}+1\right)\)
=>x+2020=0
hay x=-2020
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
a, \(\dfrac{1}{2}+\dfrac{2}{3}x=\dfrac{4}{5}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{4}{5}-\dfrac{1}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{3}{10}\\ \Rightarrow x=\dfrac{3}{10}\cdot\dfrac{3}{2}\\ \Rightarrow x=\dfrac{9}{20}\)
b, \(x+\dfrac{1}{4}=\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{4}{3}-\dfrac{1}{4}\\ \Rightarrow x=\dfrac{13}{12}\)
c, \(\dfrac{3}{5}x-\dfrac{1}{2}=-\dfrac{1}{7}\)
\(\Rightarrow\dfrac{3}{5}x=-\dfrac{1}{7}-\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{5}x=-\dfrac{9}{14}\\ \Rightarrow x=-\dfrac{9}{14}\cdot\dfrac{5}{3}\\ \Rightarrow x=\dfrac{15}{14}\)
d, \(\left|x-\dfrac{4}{5}\right|=\dfrac{3}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{4}{5}=\dfrac{3}{4}\\x-\dfrac{4}{5}=-\dfrac{3}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}+\dfrac{4}{5}\\x=-\dfrac{3}{4}+\dfrac{4}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{31}{20}\\x=\dfrac{1}{20}\end{matrix}\right.\)
e, \(lxl\) là j mk ko hiểu!
a) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\dfrac{-2}{3}x\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}x\)
\(x-\dfrac{2}{3}x=\dfrac{2}{5}-\dfrac{5}{6}\)
\(\dfrac{1}{3}x=-\dfrac{13}{30}\)
\(x=-1,3\)
b) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}-\dfrac{1}{5}x\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}-\dfrac{1}{5}x\)
\(\dfrac{7}{5}x+\dfrac{1}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{8}{5}x=-\dfrac{43}{35}\)
\(x=-\dfrac{43}{56}\)
a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
a)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\)
\(\dfrac{2}{5}+x=\dfrac{3}{12}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=\dfrac{5}{20}-\dfrac{8}{20}\)
\(x=\dfrac{-3}{20}\)
b)\(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow2x=0\) hoặc \(x-\dfrac{1}{7}=0\)
\(x=0:2\) \(x=0+\dfrac{1}{7}\)
\(x=0\) \(x=\dfrac{1}{7}\)
\(\Rightarrow x=0\) hoặc \(x=\dfrac{1}{7}\)
c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=\dfrac{8}{20}-\dfrac{15}{20}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}.\dfrac{-20}{7}\)
x= \(\dfrac{1.\left(-5\right)}{1.7}\)
\(x=\dfrac{-5}{7}\)
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
\(\Rightarrow x;1-2y\in U\left(40\right)\)
\(U\left(40\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)
Mà 1-2y lẻ nên:
\(\left\{{}\begin{matrix}1-2y=1\Rightarrow2y=0\Rightarrow y=0\\x=40\\1-2y=-1\Rightarrow2y=2\Rightarrow y=1\\x=-40\end{matrix}\right.\)
\(\left\{{}\begin{matrix}1-2y=5\Rightarrow2y=-4\Rightarrow y=-2\\x=8\\1-2y=-5\Rightarrow2y=6\Rightarrow y=3\\x=-8\end{matrix}\right.\)
b tương tự.
c) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-2>0\Rightarrow x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-2< 0\Rightarrow x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1< x< 2\Rightarrow x\in\left\{0;1\right\}\)
d tương tự
\(a,\dfrac{2}{3}-\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)-\dfrac{1}{2}\left(2x+1\right)=5\)
\(\dfrac{2}{3}-\dfrac{1}{3}x-\dfrac{1}{2}-x+\dfrac{1}{2}=5\)
\(\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}x-x=5\)
\(\dfrac{2}{3}-\dfrac{1}{3}x-x=5\)
\(\dfrac{2}{3}-\dfrac{4}{3}x=5\)
\(\dfrac{4}{3}x=\dfrac{2}{3}-5\)
\(\dfrac{4}{3}x=-\dfrac{13}{3}\)
\(x=-\dfrac{13}{3}:\dfrac{4}{3}\)
\(x=-\dfrac{13}{4}\)
Vậy...............
\(b,\left(x+\dfrac{1}{2}\right)\left(\dfrac{3}{4}-x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{3}{4}-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy................
\(c,\dfrac{2x-1}{-3+2}=0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy.............
\(a,\dfrac{2}{3}x-\dfrac{2}{5}=\dfrac{1}{2}x-\dfrac{1}{3}\\ \Rightarrow\dfrac{2}{3}x-\dfrac{1}{2}x-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\left(\dfrac{2}{3}-\dfrac{1}{2}\right)-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\dfrac{1}{6}=-\dfrac{11}{15}\\ \Rightarrow x=-\dfrac{22}{5}\\ b,\dfrac{1}{3}x+\dfrac{2}{5}.\left(x+1\right)=0\\ \Rightarrow\dfrac{1}{3}x+\left(x+1\right)=-\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{2}{5}-\left(x+1\right)\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{7}{5}-x\\ \Rightarrow\dfrac{1}{3}.2x=-\dfrac{7}{5}\\ \Rightarrow2x=-\dfrac{21}{5}\\ \Rightarrow x=-\dfrac{21}{10}.\)