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SS
23 tháng 6 2019
a, \(\frac{\left(\frac{1}{9}\right)^6\cdot\left(\frac{3}{8}\right)^7}{\left(\frac{1}{3}\right)^{13}\cdot\left(\frac{1}{2}\right)^{22}.3^6}\)
\(=\frac{\left(\frac{1}{\left(3^2\right)^6}\right)\cdot\left(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot3\right)^7}{\left(\frac{1}{3}\right)^{13}.\left(\frac{1}{2}\right)^{22}.3^6}=\frac{\frac{1}{3^{12}}\cdot\left(\frac{1}{2}\right)^{21}\cdot3^7}{\frac{1}{3^{13}}\cdot\left(\frac{1}{2}\right)^{22}.3^6}\)
\(=\frac{3}{\frac{1}{3}\cdot\frac{1}{2}}=3\div\frac{1}{6}=3.6=18\)
b, Làm tương tự nha bn
XO
4 tháng 10 2021
a) \(\left(-\frac{2}{3}\right)^2:\frac{1}{3}-\left|-1\frac{1}{2}\right|=\frac{4}{9}:\frac{1}{3}-\frac{3}{2}=\frac{4}{3}-\frac{3}{2}=-\frac{1}{6}\)
b) \(\left(\frac{1}{2}-\frac{3}{5}\right)^2+\frac{2}{3}\left|\frac{3}{4}-\frac{1}{2}\right|+2012^0=\left(-\frac{1}{10}\right)^2+\frac{2}{3},\frac{1}{4}+2012^0\)
\(=\frac{1}{100}+\frac{1}{6}+1=\frac{353}{300}\)
c) \(\left(3^2:\frac{1}{3}\right)+2^3+\frac{1}{2}+\frac{1}{4}-6=3^3+2^3+\frac{3}{4}-6=29\frac{3}{4}\)
LL
1
25 tháng 9 2015
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=5\)
\(=>\frac{1}{3}:\left(2x-1\right)=5-\frac{1}{4}=\frac{19}{4}\)
\(=>2x-1=\frac{1}{3}:\frac{19}{4}=\frac{4}{57}\)
\(=>2x=\frac{61}{57}=>x=\frac{61}{114}\)
3 tháng 3 2020
\(\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\frac{9}{4}\)
\(=>\left(\frac{1}{3}-\frac{3}{2}x\right)^2=\left(\frac{3}{2}\right)^2\)
\(=>\frac{1}{3}-\frac{3}{2}x=\frac{3}{2}\)
\(=>\frac{3}{2}x=\frac{1}{3}-\frac{3}{2}=-\frac{7}{6}\)
\(=>x=-\frac{7}{6}:\frac{3}{2}=-\frac{7}{9}\)
18 tháng 7 2016
a) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)\(5\)
=> \(\frac{2}{3}-\left(\frac{1}{3}x-\frac{1}{2}\right)-\left(x+\frac{1}{2}\right)=5\)
=>\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
=>\(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5\)
=>\(\frac{2}{3}-\frac{4}{3}x=5\)
=>\(\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)
=>\(x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{4}\)
b)\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
=>\(4x-x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)
=> \(3x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)
=>\(x=-\left(-\frac{9}{2}\right)+\frac{1}{2}=5\)
6 tháng 10 2020
a) \(\left(0,25+\frac{3}{4}:1,25+1\frac{1}{3}:2\right)\)\(=\left(\frac{1}{4}+\frac{3}{4}\right):\frac{5}{4}+\frac{4}{3}:2\)\(=\frac{4}{5}+\frac{2}{3}=\frac{22}{15}\)
b) \(2^3+3\left(\frac{-3}{2}\right)^0.\left(\frac{1}{2}\right)^2.4+\left[\left(-2\right)^2:\frac{1}{2}\right]:8\)\(=8+3.1.\frac{1}{4}.4+\left[4:\frac{1}{2}\right]:8\)
\(=8+3+8:8=\frac{19}{8}\)
\(\left(2x+1\right)^2=\dfrac{1}{9}\)
⇒ \(\left[{}\begin{matrix}2x+1=\dfrac{1}{3}\\2x+1=-\dfrac{1}{3}\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
(2x+1)\(^2\)=(1/3)\(^2\)
TH1: 2x+1=1/3 TH2:2x+1=-1/3
2x =1/3-1 2x =-1/3-1
2x =-2/3 2x =-4/3
x =-2/3:2 x =-4/3:2
x =-1/3 x =-2/3
Vậy x∈{-1/3;-2/3}