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`5/9+4/9:x=1/3`
`=>4/9:x=1/3-5/9`
`=>4/9:x=3/9-5/9`
`=>4/9:x=-2/9`
`=>x=4/9:(-2/9)`
`=>x=4/9.(-9/2)`
`=>x=-4/2`
`=>x=-2`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{-1}{4}< \dfrac{x}{24}< \dfrac{-1}{6}\\ \dfrac{-6}{24}< \dfrac{x}{24}< \dfrac{-4}{24}\\ \Rightarrow x=-5\)
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c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)
\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)
\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)
\(x=-\dfrac{15}{2}\)
d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)
\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)
\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)
A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)
\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)
\(x\)\(=\dfrac{11}{9}\)
B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)
\(x=\)\(\dfrac{-2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)(4,5-2x)*1 4/7=11/14`
`=>(4,5-2x)*11/7=11/14`
`=>4,5-2x=1/2`
`=>2x=4,5-0,5=4`
`=>x=2`
Vậy `x=2`
`b)(2,8x-32):2/3=-90`
`=>2,8x-32=-90*2/3=-60`
`=>2,8x=-28`
`=>x=-10`
Vậy `x=-10`
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\(\dfrac{2\text{x}-1}{3}=\dfrac{3\text{x}+1}{4}\)
\(\Leftrightarrow=\dfrac{4\left(2\text{x}-1\right)}{12}=\dfrac{3\left(3\text{x}+1\right)}{12}\)
\(\Leftrightarrow8\text{x}-4=9\text{x}+3\)
\(\Leftrightarrow8\text{x}-9\text{x}=3+4\)
\(\Leftrightarrow-x=7\)
\(\Leftrightarrow x=-7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
(1/3+x)2=4/9
(1/3+x)2=(2/3)2
TH1
1/3+x=2/3
x=2/3-1/3
x=1/3
TH2
1/3+x=-2/3
x=-2/3-1/3
x=-1
Vậy x bằng 1/3 và -1