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F
0
PH
0
TA
2
TM
21 tháng 7 2019
(x-3):\(\frac{-7}{8}\)=\(\frac{-5}{2}\)-x
\(\frac{x-3}{\frac{-7}{8}}\)=\(\frac{-5}{2}\)-x
x-3=(\(\frac{-5}{2}\)-x).\(\frac{-7}{8}\)
x-3=\(\frac{35}{16}+\frac{7}{8}x\)
x-3-\(\frac{7}{8}x\)=\(\frac{35}{16}\)
\(\frac{x}{8}-3=\frac{35}{16}\)
\(\frac{x}{8}=\frac{83}{16}\)=\(\frac{2x}{16}\)
x=\(\frac{83}{2}\)
N
1 tháng 8 2019
a) \(-\frac{3}{4}x+\frac{1}{6}x=1-2\frac{5}{9}\)
\(\left(-\frac{3}{4}+\frac{1}{6}\right).x=1-\frac{23}{9}\)
\(-\frac{7}{12}.x=-\frac{14}{9}\)
\(x=-\frac{14}{9}:\left(-\frac{7}{12}\right)\)
\(x=\frac{8}{3}\)
Vậy x = ...
b) \(\left|2x-\frac{3}{8}\right|+2\frac{3}{4}=3\frac{1}{16}\)
\(\left|2x-\frac{3}{8}\right|+\frac{11}{4}=\frac{49}{16}\)
\(\left|2x-\frac{3}{8}\right|=\frac{49}{16}-\frac{11}{4}\)
\(\left|2x-\frac{3}{8}\right|=\frac{5}{16}\)
\(\Rightarrow\left|2x-\frac{3}{8}\right|\in\text{{}\frac{5}{16};-\frac{5}{16}\)}
Nếu, \(2x-\frac{3}{8}=\frac{5}{16}\)
\(2x=\frac{11}{16}\)
\(x=\frac{11}{32}\)
Nếu, \(2x-\frac{3}{8}=-\frac{5}{16}\)
\(2x=\frac{1}{16}\)
\(x=\frac{1}{32}\)
Vậy \(x\in\text{{}\frac{1}{32};\frac{11}{32}\)}
NT
1
27 tháng 5 2015
\(\frac{1}{2}\left(x+1\right):\frac{3}{7}=\frac{32}{135}\)
\(\frac{1}{2}\left(x+1\right)=\frac{32}{135}.\frac{3}{7}\)
\(\frac{1}{2}\left(x+1\right)=\frac{32}{315}\)
\(x+1=\frac{32}{315}:\frac{1}{2}\)
\(x+1=\frac{64}{315}\)
\(x=\frac{64}{315}-1=-\frac{251}{315}\)
\(\left(x-3\right):\left[\frac{1}{2}.\left(-1\frac{3}{4}\right)\right]=-2\frac{1}{2}\)
\(\left(x-3\right):\left[\frac{1}{2}.\left(\frac{-7}{4}\right)\right]=\frac{-5}{2}\)
\(\left(x-3\right):\frac{-7}{8}=\frac{-5}{2}\)
\(\left(x-3\right)=\frac{-5}{2}.\frac{-7}{8}\)
\(\left(x-3\right)=\frac{35}{16}\)
\(x=\frac{35}{16}+3=\frac{83}{16}\)
\(\left(x-3\right):\left[\frac{1}{2}.\left(-1\frac{3}{4}\right)\right]=-2\frac{1}{2}\)
\(\left(x-3\right):\left(\frac{1}{2}.\frac{-7}{4}\right)=\frac{-5}{2}\)
\(\left(x-3\right):\frac{-7}{8}=\frac{-5}{2}\)
\(x-3=\frac{-5}{2}.\frac{-7}{8}\)
\(x-3=\frac{35}{16}\)
\(x=\frac{35}{16}+3\)
\(x=\frac{35+48}{16}=\frac{83}{16}\)