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PH
0
7 tháng 10 2017
\(5+x-3=5-\left(x+4\right)\)
\(5+x-3=5-x-4\)
\(x+x=5-4-5+3\)
\(2x=-1\)
\(x=\frac{-1}{2}\)
vay \(x=\frac{-1}{2}\)
\(x-\left(5.\left|-7\right|+3\right)=\left|-8\right|+6-2\)
\(x-\left(5.7+3\right)=8+6-2\)
\(x-38=12\)
\(x=12+38\)
\(x=50\)
vay \(x=50\)
\(\left|x-3\right|+7=0\)
\(\left|x-3\right|=-7\)( ko ton tai)
\(12-3\left|x-1\right|=6\)
\(3\left|x-1\right|=6\)
\(\left|x-1\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
vay \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
toi nay mk se lay y kien cua cac ban , giup mk nha
CV
12 tháng 7 2017
b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
15 tháng 4 2019
\(a,\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\cdot(-2)^2\)
\(=\frac{6}{7}+\frac{5}{8}:\frac{5}{1}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{5}{8}\cdot\frac{1}{5}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3\cdot4}{16}\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3\cdot1}{4}\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}=\frac{48+7-42}{56}=\frac{13}{56}\)
\(b,\frac{2}{3}+\frac{1}{3}\cdot\left[\frac{-2}{3}+\frac{5}{6}\right]:\frac{2}{3}\)
\(=\frac{2}{3}+\frac{1}{3}\cdot\left[\frac{-4+5}{6}\right]:\frac{2}{3}\)
\(=\frac{2}{3}+\frac{1}{3}\cdot\frac{1}{6}:\frac{2}{3}=\frac{2}{3}+\frac{1}{3}\cdot\frac{1}{6}\cdot\frac{3}{2}=\frac{2}{3}+\frac{1}{12}=\frac{8}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}\)
c, Xem lại đề
d, \(\frac{-3}{5}+\left[\frac{-2}{5}-99\right]\)
\(=\frac{-3}{5}+\frac{-497}{5}=\frac{-500}{5}=-100\)
b, Tìm x
\(\left[\frac{2}{11}+\frac{1}{3}\right]\cdot x=\left[\frac{1}{7}-\frac{1}{8}\right]\cdot56\)
\(\Rightarrow\left[\frac{2}{11}+\frac{1}{3}\right]\cdot x=\left[\frac{8}{56}-\frac{7}{56}\right]\cdot56\)
\(\Rightarrow\left[\frac{6}{33}+\frac{11}{33}\right]\cdot x=1\)
\(\Rightarrow\frac{17}{33}\cdot x=1\)
\(\Rightarrow x=1:\frac{17}{33}=1\cdot\frac{33}{17}=\frac{33}{17}\)
25 tháng 6 2018
\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
18 tháng 2 2021
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