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LH
13 tháng 8 2020
a) \(2x\left(x-\frac{1}{7}\right)=0\)
⇒\(\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)
Vậy \(x=0;x=\frac{1}{7}\)
b) \(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{25}\\ \left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{25}\\ \left(\frac{5}{10}+\frac{6}{10}\right)x=\frac{-33}{25}\\ \frac{11}{10}x=\frac{-33}{25}\\ x=\frac{-33}{25}:\frac{11}{10}\\ x=\frac{-33.10}{25.11}\\ x=\frac{-6}{5}\)
Vậy x = \(\frac{-6}{5}\)
c) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\\ \Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=\frac{4}{9}\\\frac{-3}{7}:x=\frac{-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{4}{9}:\frac{2}{3}=\frac{4.3}{9.2}=\frac{2}{3}\\x=\frac{-3}{7}:\frac{-1}{2}=\frac{-3.2}{7.\left(-1\right)}=\frac{6}{7}\end{matrix}\right.\)
TG
13 tháng 8 2020
a) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0:2=0\\x=0+\frac{1}{7}=\frac{1}{7}\end{matrix}\right.\)
b) \(\frac{1}{2}x+\frac{3}{5}x=-\frac{33}{25}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{5}\right)=-\frac{33}{25}\)
\(\Rightarrow x\frac{11}{10}=-\frac{33}{25}\)
\(\Rightarrow x=\left(-\frac{33}{25}\right):\frac{11}{10}=-\frac{33}{25}.\frac{10}{11}=-\frac{6}{5}\)
c) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=0+\frac{4}{9}=\frac{4}{9}\\-\frac{3}{7}:x=0-\frac{1}{2}=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{4}{9}:\frac{2}{3}=\frac{4}{9}.\frac{3}{2}=\frac{2}{3}\\x=\left(-\frac{3}{7}\right):\frac{-1}{2}=\left(-\frac{3}{7}\right).\left(-2\right)=\frac{6}{7}\end{matrix}\right.\)
12 tháng 12 2018
a)\(0,45-\left|1,3-x\right|=0\)
\(\Leftrightarrow\left|1,3-x\right|=0,45-0\)
\(\Leftrightarrow\hept{\begin{cases}1,3-x=0,45\\1,3-x=-0,45\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=1,3-0,45\\x=1,3+0,45\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0,85\\x=1,75\end{cases}}\)
Vậy x = 0,85 ; x = 1,75
b) \(\left|3x-5\right|-\frac{1}{7}=\frac{1}{3}\)
\(\Leftrightarrow\left|3x-5\right|=\frac{1}{3}+\frac{1}{7}\)
\(\Leftrightarrow\left|3x-5\right|=\frac{10}{21}\)
\(\Leftrightarrow\hept{\begin{cases}3x-5=\frac{10}{21}\\3x-5=-\frac{10}{21}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}3x=\frac{10}{21}+5\\3x=-\frac{10}{21}+5\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}3x=\frac{115}{21}\\3x=\frac{95}{21}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{115}{63}\\x=\frac{95}{63}\end{cases}}\)
Vậy x = .........................
XO
4 tháng 10 2020
\(\left|x-\frac{2}{5}\right|+\frac{1}{2}=\frac{3}{4}\)
=> \(\left|x-\frac{2}{5}\right|=\frac{1}{4}\)
=> \(\orbr{\begin{cases}x-\frac{2}{5}=\frac{1}{4}\\x-\frac{2}{5}=-\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{20}\\x=\frac{3}{20}\end{cases}}\)
b) \(\left|5-3x\right|+\frac{2}{3}=\frac{11}{6}\)
=> \(\left|5-3x\right|=\frac{7}{6}\)
=> \(\orbr{\begin{cases}5-3x=\frac{7}{6}\\5-3x=-\frac{7}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}3x=\frac{23}{6}\\3x=\frac{37}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{23}{18}\\x=\frac{37}{18}\end{cases}}\)
c) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=6^2\)
=> \(\left|x+\frac{3}{4}\right|=36,5\)
=> \(\orbr{\begin{cases}x+\frac{3}{4}=36,5\\x+\frac{3}{4}=-36,5\end{cases}}\Rightarrow\orbr{\begin{cases}x=35,75\\x=-37,25\end{cases}}\)
4 tháng 10 2020
a) \(\left|x-\frac{2}{5}\right|+\frac{1}{2}=\frac{3}{4}\Leftrightarrow\left|x-\frac{2}{5}\right|=\frac{1}{4}\)\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{5}=\frac{1}{4}\\x-\frac{2}{5}=\frac{-1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{20}\\x=\frac{3}{20}\end{cases}}}\)
b) \(\left|5-3x\right|+\frac{2}{3}=\frac{11}{6}\Leftrightarrow\left|5-3x\right|=\frac{7}{6}\)\(\Leftrightarrow\orbr{\begin{cases}5-3x=\frac{7}{6}\\5-3x=\frac{-7}{6}\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=\frac{23}{6}\\3x=\frac{37}{6}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{23}{18}\\x=\frac{37}{18}\end{cases}}}\)
c) \(\frac{1}{5}-\left|\frac{1}{5}-x\right|=\frac{1}{5}\Leftrightarrow\left|\frac{1}{5}-x\right|=0\Leftrightarrow x=\frac{1}{5}\)
BN
5
CC
8 tháng 12 2019
\(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{10}\)\(\Leftrightarrow\left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{10}\)
\(\Leftrightarrow\frac{11}{10}x=\frac{-33}{10}\)\(\Leftrightarrow x=\frac{-33}{10}:\frac{11}{10}=\frac{-33}{10}.\frac{10}{11}=-3\)
Vậy \(x=-3\)
8 tháng 12 2019
\(\frac{1}{2}\cdot x+\frac{3}{5}\cdot x=-\frac{33}{10}\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)\cdot x=-\frac{33}{10}\)
\(\left(\frac{5}{10}+\frac{6}{10}\right)\cdot x=-\frac{33}{10}\)
\(\frac{11}{10}\cdot x=-\frac{33}{10}\)
\(x=-\frac{33}{10}:\frac{11}{10}=-\frac{33}{10}\cdot\frac{10}{11}\)
\(x=-\frac{33}{11}=-3\)
\(\frac{1}{2}x+\frac{3}{5}x=\frac{-33}{25}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-33}{25}\)
\(\Rightarrow\frac{11}{10}x=\frac{-33}{25}\)
\(\Rightarrow x=\frac{-33}{25}\div\frac{11}{10}\)
\(\Rightarrow x=\frac{-6}{5}\)
Vậy \(x=\frac{-6}{5}.\)
\(\frac{1}{2}.x+\frac{3}{5}.x=\frac{-33}{25}\)
\(\Leftrightarrow x.\left(\frac{1}{2}+\frac{3}{5}\right)=\frac{-33}{25}\)
\(\Leftrightarrow x.\left(\frac{5}{10}+\frac{6}{10}\right)=\frac{-33}{25}\)
\(\Leftrightarrow x.\frac{11}{10}=\frac{-33}{25}\)
\(\Leftrightarrow x=\frac{-33}{25}:\frac{11}{10}=\frac{-33}{25}.\frac{10}{11}\)
\(\Leftrightarrow x=\frac{-330}{275}=\frac{-6}{5}\)