TÍNH X
(0,6x - \(\frac{1}{2}\)).\(\frac{3}{4}\)-(-1)=1/3
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\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}+1=\frac{1}{3}\)
\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}=\frac{1}{3}-1\)
\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}=\frac{-2}{3}\)
\(-0,6x-\frac{1}{2}=-\frac{2}{3}:\frac{3}{4}\)
\(-0,6x=-\frac{8}{9}\)
\(x=\frac{-8}{9}:\left(-0,6\right)\)
\(x=\frac{40}{27}\)
c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};3\right\}\)
e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
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1/1+2 + 1/1+2+3 +1/1+2+3+4 +...+1/1+2+3+...+50
Ta có 2/2(1+2)+2/2(1+2+3)+...+2/2(1+2+...+50)
=2/6+2/12+2/20+...+2/2550
=2/2.3+2/3.4+...+2/50.51
=2(1/2.3+1/3.4+...+1/50.51)
=2(1/1-1/2+1/2-...+1/50-1/51)
=2.(1-1/51)
=2.50/51=100/51
\(3.\left(x-\frac{1}{5}\right)-7.\left(\frac{5}{14}-3\right)=20\)
\(3.\left(x-\frac{1}{5}\right)-7.\frac{-37}{14}=20\)
\(3.\left(x-\frac{1}{5}\right)-\frac{-37}{2}=20\)
\(3.\left(x-\frac{1}{5}\right)=20+\frac{-37}{2}\)
\(3.\left(x-\frac{1}{5}\right)=\frac{3}{2}\)
\(x-\frac{1}{5}=\frac{3}{2}:3\)
\(x-\frac{1}{5}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{1}{5}\)
\(x=\frac{7}{10}\)
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