giải chi tiết các bước
\((\dfrac{1}{5}+\dfrac{-8}{19}-\dfrac{-7}{30})-(\dfrac{11}{19}-\dfrac{4}{5})=?\)
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Ta có:
\(2x+33=\left(-11\right)-x\)
\(\Leftrightarrow2x+33=\left(-11\right)+\left(-x\right)\)
\(\Leftrightarrow2x+33=-\left(11+x\right)\)
\(\Leftrightarrow2x=-\left(11+x\right)-33\)
\(\Leftrightarrow2x=-\left(11+x\right)+\left(-33\right)\)
\(\Leftrightarrow2x=-\left(11+33+x\right)\)
\(\Leftrightarrow2x=-\left(44+x\right)\)
\(\Leftrightarrow2x=2\cdot\frac{-\left(44+x\right)}{2}\)
\(\Leftrightarrow2x=2\cdot\left[\left(-44\right):2\right]+\left[\left(-x\right):2\right]\)
\(\Leftrightarrow2x=2\cdot\left(-22\right)+\frac{-x}{2}\)
\(\Leftrightarrow x=\left(-22\right)+\frac{-x}{2}\)
\(\Leftrightarrow\left(-22\right)+\frac{-x}{2}=x\)
\(\Leftrightarrow\frac{-x}{2}=x-22\)
\(\Leftrightarrow\frac{-x}{2}=-22\)
\(\Leftrightarrow-x=\left(-22\right):2\)
\(\Leftrightarrow-x=-11\)
\(\Leftrightarrow x=11\)
\(\left(2x+1\right)^2-9=0\)
\(\Leftrightarrow\left(2x+1\right)^2=9\)
\(\Leftrightarrow\left(2x+1\right)^2=3^2\)
\(\Leftrightarrow\left(2x+1\right)=3\)
\(\Leftrightarrow2x=3-1\)
\(\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\)
Vậy x = 1
Ta có:
\(\left(2x+1\right)^2-9=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0+9\)
\(\Leftrightarrow\left(2x+1\right)^2=9\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x+1\right)^2=3^2\\\left(2x+1\right)^2=-3^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=3\\2x+1=-3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x=3-1\\2x=\left(-3\right)-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x=2\\2x=-4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2:2\\x=\left(-4\right):2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\x=-2\end{cases}}\)
=\(\frac{1}{5}+\frac{-8}{19}-\frac{-7}{30}-\frac{11}{19}+\)\(\frac{4}{5}\)
\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-8}{19}-\frac{11}{19}\right)\)\(+\frac{7}{30}\)
\(=1+\left(-1\right)+\frac{7}{30}\)
\(=\frac{7}{30}\).