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![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>2x-1=-2
=>2x=-1
hay x=-1/2
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}\)
c: x/8=9/4
nên x/8=18/8
hay x=18
d: \(\Leftrightarrow\left(x-3\right)^2=36\)
=>x-3=6 hoặc x-3=-6
=>x=9 hoặc x=-3
e: =>-1,7x=6,12
hay x=-3,6
h: =>x-3,4=27,6
hay x=31
a) \(\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}\)
\(\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)\)
\(\dfrac{1}{3}=-1\left(2x-1\right)\div6\)
\(\dfrac{1}{3}=-2x+1\div6\)
\(x=-\dfrac{1}{2}\)
b) \(\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0\)
\(TH1:3x+2=0\)
\(3x=0-2\)
\(3x=-2\)
\(x=\dfrac{-2}{3}\)
\(TH2:\left(-\dfrac{2}{5}x-7\right)=0\)
\(\left(\dfrac{-2}{5}x-7\right)=0\)
\(\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0\)
\(\left(\dfrac{-2x-35}{5}\right)=0\)
\(-2x-35=0\)
\(-2x=0+35\)
\(x=-\dfrac{35}{2}\)
c) \(\dfrac{x}{8}=\dfrac{9}{4}\)
\(\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18\)
\(x=18\)
d) \(\dfrac{x-3}{2}=\dfrac{18}{x-3}\)
\(x-3=18+2\)
\(x=20-3\)
\(x=17\)
e) \(4,5x-6,2x=6,12\)
\(\dfrac{9x}{2}-6,2.x=6,12\)
\(\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12\)
\(\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12\)
\(\dfrac{45x-62x}{10}=6.12\)
\(=-17x\div10=6.12\)
\(-17x=10.6.12\)
\(x=-3,6\)
h) \(11,4-\left(x-3,4\right)=-16,2\)
\(x-3,4=-16,2+11,4\)
\(x-3,4=-4,8\)
\(x=-1,4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x-\dfrac{5}{8}\right).\dfrac{5}{8}=-\dfrac{15}{36}\)
\(\left(x-\dfrac{5}{8}\right)=-\dfrac{15}{36}\div\dfrac{5}{8}\)
\(x-\dfrac{5}{8}=-\dfrac{2}{3}\)
\(x=-\dfrac{2}{3}+\dfrac{5}{8}\)
\(x=-\dfrac{1}{24}\)
\(b,\left(x-\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x-\dfrac{1}{3}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}+\dfrac{1}{3}\)
\(x=\dfrac{7}{6}\)
\(a,\left(x-\dfrac{5}{8}\right)\cdot\dfrac{8}{18}=-\dfrac{15}{16}\\ x-\dfrac{5}{8}=-\dfrac{15}{36}:\dfrac{8}{18}\\ x-\dfrac{5}{8}=-\dfrac{15}{16}\\ x=-\dfrac{15}{16}+\dfrac{5}{8}\\ x=-\dfrac{15}{16}+\dfrac{10}{16}\\ x=-\dfrac{5}{16}\\ b,x-\dfrac{1}{3}=\dfrac{5}{6}\\ x=\dfrac{5}{6}+\dfrac{1}{3}\\ x=\dfrac{5}{6}+\dfrac{2}{6}\\ x=\dfrac{7}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A -\(\dfrac{24}{25}\)
B -\(\dfrac{5}{21}\)
C -\(\dfrac{24}{47}\)
D -\(\dfrac{19}{42}\)
tick cho mk
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\\ =>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\\ =>2x-\dfrac{1}{2}=\dfrac{10}{12}-\dfrac{15}{12}\\ =>2x-\dfrac{1}{2}=-\dfrac{5}{12}\\ =>2x=-\dfrac{5}{12}+\dfrac{1}{2}\\ =>2x=-\dfrac{5}{12}+\dfrac{6}{12}\\ =>2x=\dfrac{1}{12}\\ =>x=\dfrac{1}{12}:2\\ =>x=\dfrac{1}{12}\cdot\dfrac{1}{2}\\ =>x=\dfrac{1}{24}\)
__
\(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{12}{8}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{7}{8}\\ =>x=\dfrac{7}{8}-\dfrac{1}{4}\\ =>x=\dfrac{7}{8}-\dfrac{2}{8}\\ =>x=\dfrac{5}{8}\)
__
\(\dfrac{x}{3}=\dfrac{12}{x}\\ =>x^2=3\cdot12\\ =>x^2=36\\ =>x^2=6^2\\ =>x=\pm6\)
Tìm x:
a) \(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\)
\(=>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\)
\(=>2x-\dfrac{1}{2}=\dfrac{-5}{12}\)
\(=>2x=\dfrac{-5}{12}+\dfrac{1}{2}\)
\(=>2x=\dfrac{1}{12}\)
\(=>x=\dfrac{1}{12}:2\)
\(=>x=\dfrac{1}{24}\)
b) \(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{7}{8}\)
\(=>x=\dfrac{7}{8}-\dfrac{1}{4}\)
\(=>x=\dfrac{5}{8}\)
c) \(\dfrac{x}{3}=\dfrac{12}{x}\)
Ta có: \(x.x=3.12\)
\(\Rightarrow x^2=36\)
Vậy x = 6 hoặc x = -6
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>6/x=x/24
=>x^2=144
=>x=12 hoặc x=-12
b: =>x(1-7/12+3/8)=5/24
=>x*19/24=5/24
=>x=5/24:19/24=5/19
c: =>(x-1/3)^2=1+3/4+1/2=9/4
=>x-1/3=3/2 hoặc x-1/3=-3/2
=>x=11/6 hoặc x=-7/6
d: =>(x-3)^2=16
=>x-3=4 hoặc x-3=-4
=>x=-1 hoặc x=7
e: =>9/x=-1/3
=>x=-27
f: =>x-1/2=0 hoặc -x/2-3=0
=>x=1/2 hoặc x=-6
\(TH1:x\ge0\)
\(\Rightarrow x\left(1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)
\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{1-\dfrac{5}{6}}=5\left(TM\right)\)
\(TH2:x< 0\)
\(\Rightarrow x\left(-1-\dfrac{5}{6}\right)=\dfrac{4}{9}.\dfrac{15}{8}\)
\(\Rightarrow x=\dfrac{\dfrac{4}{9}.\dfrac{15}{8}}{-1-\dfrac{5}{6}}=-\dfrac{5}{11}\left(TM\right)\)
Vậy ...
Giải:
\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\)
\(TH1:x\ge0\)
\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\)
\(x-\dfrac{5}{6}.x=\dfrac{5}{6}\)
\(x.\left(1-\dfrac{5}{6}\right)=\dfrac{5}{6}\)
\(x.\dfrac{1}{6}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}:\dfrac{1}{6}\)
\(x=5\)
\(TH2:x\le0\)
\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\)
\(-x-\dfrac{5}{6}.x=\dfrac{5}{6}\)
\(x.\left(-1-\dfrac{5}{6}\right)=\dfrac{5}{6}\)
\(x.\dfrac{-11}{6}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}:\dfrac{-11}{6}\)
\(x=\dfrac{-5}{11}\)
Vậy \(x\in\left\{\dfrac{-5}{11};5\right\}\)