Tim x: \(\frac{\frac{-2}{3}}{x} = \dfrac{9}{10}\)
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VT
2
TM
27 tháng 11 2016
Đặt \(\frac{x-2}{6}=\frac{y+3}{9}=\frac{z-7}{10}=k\Rightarrow\hept{\begin{cases}x=6k+2\\y=9k-3\\z=10k+7\end{cases}}\)
Theo đề bài: x+y+z=106
<=>\(6k+2+9k-3+10k+7=106\)
<=>\(25k+6=106\)
<=> 25k = 100
<=> k = 4
=> \(\hept{\begin{cases}x=6.4+2=26\\y=9.4-3=33\\z=10.4+7=47\end{cases}}\)
Vậy .........................
1 tháng 6 2020
\(\frac{8}{23}\cdot\frac{46}{24}-x=\frac{1}{3}\)
=> \(\frac{2}{3}-x=\frac{1}{3}\)
=> \(x=\frac{1}{3}\)
\(\frac{10}{12}\div x=\frac{28}{9}\cdot\frac{3}{56}\)
=> \(\frac{10}{12}\div x=\frac{1}{6}\)
=> \(x=\frac{60}{12}=5\)
\(\frac{x-12}{4}=\frac{1}{2}\)
=> \(\left(x-12\right)\cdot2=4\cdot1\)
=> \(2x-24=4\)
=> \(2x=28\)
=> \(x=14\)
8 tháng 7 2017
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
26 tháng 2 2020
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\left(x\ne1\right)\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x}{x^2+x+1}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\left(x^2+x+1-3x^2-2x^2+2x\right)=0\)
\(\Leftrightarrow-4x^2+3x+1=0\left(\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\ne0\right)\)
\(\Leftrightarrow-4x^2+4x-x+1=0\)
\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-4x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\-4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\-4x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\left(loại\right)\\x=\frac{-1}{4}\end{cases}}}\)
Vậy \(x=\frac{-1}{4}\)
MT
10 tháng 6 2020
c) \(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{15}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}.\frac{27}{8}\)
\(2+\frac{3}{4}x=\frac{21}{8}\)
\(\frac{3}{4}x=\frac{21}{8}-2\)
\(\frac{3}{4}x=\frac{21}{8}-\frac{16}{8}\)
\(\frac{3}{4}x=\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{3}{4}\)
\(x=\frac{5}{8}.\frac{4}{3}\)
\(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}\).
d) \(\left|x-\frac{1}{3}\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x-\frac{1}{3}\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x-\frac{1}{3}\right|=\frac{20}{12}+\frac{9}{12}\)
\(\left|x-\frac{1}{3}\right|=\frac{29}{12}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{29}{12}\\x-\frac{1}{3}=-\frac{29}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=-\frac{25}{12}\end{cases}}\)
Vậy \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).
\(\dfrac{3}{x}=2:\dfrac{9}{10}\Rightarrow\dfrac{3}{x}=\dfrac{20}{9}\Rightarrow x=3:\dfrac{20}{9}=\dfrac{27}{20}\)
\(\dfrac{-2}{\dfrac{3}{x}}=\dfrac{9}{10}\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{9}{10}\Leftrightarrow x=\dfrac{-20}{27}\)