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29 tháng 6 2017
a) \(\frac{2}{3}x\times\frac{1}{2}=\frac{1}{10}\Rightarrow\frac{2}{3}x=\frac{1}{5}\Rightarrow x=\frac{3}{10}\)
TG
12 tháng 4 2020
a/ \(2x+\frac{1}{7}=\frac{1}{3}\)
=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)
=> \(2x=\frac{4}{21}\)
=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)
b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)
=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)
=> \(x-\frac{1}{2}=\frac{4}{27}\)
=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)
c/ \(\left(x-5\right)^2+4=68\)
=> \(\left(x-5\right)^2=68-4=64\)
=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)
d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)
=> \(\left[{}\begin{matrix}\left|x\right|-\frac{1}{2}=0\\2x+\frac{3}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left|x\right|=0+\frac{1}{2}=\frac{1}{2}\\2x=0-\frac{3}{2}=-\frac{3}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)
e) \(5x+2=3x+8\)
=> \(5x-3x=8-2=6\)
=> \(2x=6\)
=> \(x=6:2=3\)
f/ \(26-\left(5-2x\right)=27\)
=> \(5-2x=26-27=-1\)
=> \(2x=5-\left(-1\right)=5+1=6\)
=> \(x=6:2=3\)
g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)
=> \(4x-8-2x+6=4\)
=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)
=> \(2x+-2=4\)
=> \(2x=4+2=6\)
=> \(x=6:2=3\)
h/ \(\left(x+3\right)^3:3-1=-10\)
=> \(\left(x+3\right)^3:3=-10+1=-9\)
=> \(\left(x+3\right)^3=-9.3=-27\)
=> \(x+3=-3\)
=> \(x=-3-3=-6\)
NT
0
TN
0
18 tháng 6 2022
Câu 1:
a: \(\Leftrightarrow43-\left|x\right|=17-45=-28\)
\(\Leftrightarrow\left|x\right|=71\)
hay \(x\in\left\{71;-71\right\}\)
b: \(-12\left(x+5\right)+7\left(3-x\right)=5\)
=>-12x-60+21-7x=5
=>-19x-39=5
=>-19x=44
hay x=-44/19
c: \(\Leftrightarrow2x^2=29+3=32\)
=>x=4 hoặc x=-4
10 tháng 5 2020
a, \(\frac{3}{5}+\frac{-4}{15}=\frac{9}{15}-\frac{4}{15}=\frac{5}{15}=\frac{1}{3}\)
b, \(\frac{-1}{3}+\frac{2}{5}+\frac{2}{15}=\frac{-5}{15}+\frac{6}{15}+\frac{2}{15}=\frac{3}{15}=\frac{1}{5}\)
c, \(\frac{-3}{5}+\frac{7}{21}+\frac{-4}{5}+\frac{7}{5}=\frac{-3}{5}+\frac{1}{3}+\frac{-4}{5}+\frac{7}{5}=\left(\frac{-3}{5}+\frac{-4}{5}+\frac{7}{5}\right)+\frac{1}{3}=\frac{1}{3}\)
d, \(\frac{2}{7}+\frac{1}{9}+\frac{3}{7}+\frac{5}{9}+\frac{-5}{6}=\left(\frac{2}{7}+\frac{3}{7}\right)+\left(\frac{1}{9}+\frac{5}{9}\right)+\frac{-5}{6}=\frac{5}{7}+\frac{6}{9}+\frac{-5}{6}=\frac{90}{126}+\frac{84}{126}+\frac{-105}{126}=\frac{69}{126}=\frac{23}{42}\)
e, \(\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}=\left(\frac{-5}{7}+\frac{-2}{7}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{-1}{5}=\left(-1\right)+1+\frac{-1}{5}=\frac{-1}{5}\)
f, \(\frac{-3}{31}+\frac{-6}{17}+\frac{1}{25}+\frac{-28}{31}+\frac{-1}{17}+\frac{-1}{5}=\left(\frac{-3}{31}+\frac{-28}{31}\right)+\left(\frac{-6}{17}+\frac{-1}{17}\right)+\left(\frac{1}{25}+\frac{-1}{5}\right)=\left(-1\right)+\frac{-7}{17}+\frac{-4}{25}=\frac{-425}{425}+\frac{-175}{425}+\frac{-68}{425}=\frac{-668}{425}\)
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