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a) \(0,5\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x+\left(-\dfrac{4}{5}\right)\)
\(\Rightarrow\dfrac{1}{2}\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x-\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{6}-\dfrac{1}{2}=x-\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{2}{3}=x-\dfrac{4}{5}\)
\(\Leftrightarrow15x-20=30x-24\)
\(\Leftrightarrow15x-30x=-24+20\)
\(\Leftrightarrow-15x=-4\)
\(\Rightarrow x=\dfrac{4}{15}\)
Vậy \(x=\dfrac{4}{15}\)
b) \(\dfrac{2}{3}\cdot\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)+\dfrac{1}{2}=\dfrac{2}{3}-x\)
\(\Rightarrow\dfrac{1}{3}x-\dfrac{2}{9}+\dfrac{1}{2}=\dfrac{2}{3}-x\)
\(\Leftrightarrow\dfrac{1}{3}x+\dfrac{5}{18}=\dfrac{2}{3}-x\)
\(\Leftrightarrow6x+5=12-18x\)
\(\Leftrightarrow6x+18x=12-5\)
\(\Leftrightarrow24x=7\)
\(\Rightarrow x=\dfrac{7}{24}\)
Vậy \(x=\dfrac{7}{24}\)
\(a,0,5\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x+\dfrac{-4}{5}\\ \dfrac{1}{2}\cdot\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}=x+\dfrac{-4}{5}\\ \dfrac{1}{2}x-\dfrac{1}{6}-\dfrac{1}{2}=x+\dfrac{-4}{5}\\ \dfrac{1}{2}x-x=\dfrac{-4}{5}+\dfrac{1}{6}+\dfrac{1}{2}\\ x\left(\dfrac{1}{2}-1\right)=\dfrac{-24}{30}+\dfrac{5}{30}+\dfrac{15}{30}\\ \dfrac{-1}{2}x=\dfrac{-2}{15}\\ x=\dfrac{-2}{15}:\dfrac{-1}{2}\\ x=\dfrac{-2}{15}\cdot\left(-2\right)\\ x=\dfrac{4}{15}\)
\(b,\dfrac{2}{3}\cdot\left(\dfrac{1}{2}\cdot x-\dfrac{1}{3}\right)+\dfrac{1}{2}=\dfrac{2}{3}-x\\ \dfrac{1}{3}x-\dfrac{2}{9}+\dfrac{1}{2}=\dfrac{2}{3}-x\\ \dfrac{1}{3}x+x=\dfrac{2}{3}+\dfrac{2}{9}-\dfrac{1}{2}\\ x\left(\dfrac{1}{3}+1\right)=\dfrac{12}{18}+\dfrac{4}{18}-\dfrac{9}{18}\\ \dfrac{4}{3}x=\dfrac{7}{18}\\ x=\dfrac{7}{18}:\dfrac{4}{3}\\ x=\dfrac{7}{18}\cdot\dfrac{3}{4}\\ x=\dfrac{7}{24}\)
a: =>|x-1/4|=3/4
=>x-1/4=3/4 hoặc x-1/4=-3/4
=>x=1 hoặc x=-1/2
b: \(\left|x+\dfrac{1}{2}\right|=\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{2-9}{4}=-\dfrac{7}{4}\)(vô lý)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{4}{3};-6\right\}\)
e: =>|3/2-x|=0
=>3/2-x=0
hay x=3/2
a, \(\left|x+1,2\right|=0,5\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1,2=0,5\\x+1,2=-0,5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-0,7\\x=-1,7\end{matrix}\right.\)
Vậy ....
b, \(\left|x-\dfrac{1}{2}\right|+\dfrac{5}{6}=1\dfrac{1}{2}\)
\(\Leftrightarrow\left|x-\dfrac{1}{2}\right|=1\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{2}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{2}{3}\\x-\dfrac{1}{2}=\dfrac{-2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{-1}{6}\end{matrix}\right.\)
Vậy .....
c, \(\left|x-\dfrac{1}{2}\right|+\dfrac{4}{5}=\left|-3,2+\dfrac{2}{5}\right|\)
\(\left|x-\dfrac{1}{2}\right|+\dfrac{4}{5}=\left|-2,8\right|\)
\(\left|x-\dfrac{1}{2}\right|+\dfrac{4}{5}=2,8\)
\(\left|x-\dfrac{1}{2}\right|=2,8-\dfrac{4}{5}\)
\(\left|x-\dfrac{1}{2}\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=2\\x-\dfrac{1}{2}=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-8}{5}\end{matrix}\right.\)
Vậy ...
dễ lắm
1) \(\left|0,5-\dfrac{1}{3}+x\right|=-\left|y\right|-\dfrac{1}{4}\)
\(\Leftrightarrow\left|0,5-\dfrac{1}{3}+x\right|=-\left(\left|y\right|+\dfrac{1}{4}\right)\)
Vì \(\left|y\right|\ge0\forall y\Rightarrow-\left(\left|y\right|+\dfrac{1}{3}\right)< 0\forall y\)
VT>0; VP<0=> PT vô nghiệm
2
Dấu bằng xảy ra \(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+\dfrac{13}{7}=0\\z-2008=0\end{matrix}\right.\)\(\Leftrightarrow z=2008;x=-\dfrac{13}{7}\)
MÌNH GHI KẾT QUẢ THÔI NHÉ
A,2
b,\(\dfrac{-2}{3}\)hoặc2
c,\(\dfrac{13}{15}\)hoặc\(\dfrac{17}{-15}\)
\(\Leftrightarrow\left(x-0.5\right)\cdot\dfrac{-4}{x-0.5}=-1\cdot\left(-4\right)\)
=>-4=4(loại)