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\(\frac{2}{3}-\frac{1}{3}.\frac{x-3}{2}-\frac{1}{2}.2.x+1=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{x-3}{3.2}-\frac{2.x}{2}+1=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{x-3}{6}-x+1=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{x-3}{6}-x=4\)
\(\Leftrightarrow\frac{4}{6}-\frac{x-3}{6}-\frac{6x}{6}=4\)
\(\Leftrightarrow\frac{4-\left(x-3\right)-6x}{6}=4\)
\(\Leftrightarrow\frac{4-x+3+6x}{6}=4\)
\(\Leftrightarrow\frac{4+3-x+6x}{6}=\frac{4}{1}\)
\(\Leftrightarrow\frac{7+5x}{6}=\frac{4}{1}\)
\(\Leftrightarrow7+5x=4.6\)
\(\Leftrightarrow7+5x=24\)
\(\Leftrightarrow5x=24-7\)
\(\Leftrightarrow5x=17\)
\(\Leftrightarrow x=\frac{17}{5}\)
Vậy \(x=\frac{17}{5}\)
Chúc bạn học tốt
a)\(\frac{1}{4}.x=-\frac{1}{3}\)
\(x=-\frac{1}{3}:\frac{1}{4}\)
\(x=-\frac{4}{3}\)
b)\(-\frac{3}{7}+x=\frac{5}{8}\)
\(\text{ }x=\frac{5}{8}-\left(-\frac{3}{7}\right)\)
\(x=\frac{59}{56}\)
c)\(\frac{16}{2^x}=2\)
\(2^x=\frac{16}{2}\)
\(2^x=8\)
\(\Rightarrow2^x=2^3\)
vậy x=3
1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)
\(\)\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\left(\dfrac{1}{3}\right)^5\)
⇒\(\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\\\dfrac{2}{3}x-\dfrac{1}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
⇒\(\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{2}{3}\\\dfrac{2}{3}x=0\end{matrix}\right.\)
⇒\(\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
Đề như thế này đúng ko?: \(3\frac{1}{3}:2\frac{1}{2}< x< 7\frac{2}{3}.\frac{3}{7}+\frac{5}{2}\)
5 . y . \(\frac{1}{2}\). x3y(\(\frac{-1}{3}\).x2.y)3= \(\frac{5}{2}\)x3y2 \(\frac{-1}{27}\) x6y3= \(\frac{-5}{54}\)x9y5
Hệ số \(\frac{-5}{54}\)
Phần biến : x9y5
Bậc : 14
Chúc bạn học tốt !!!
(\(x\) + \(\dfrac{1}{2}\))2 = \(\dfrac{1}{16}\)
\(\left[{}\begin{matrix}x+\dfrac{1}{2}=-\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{4}-\dfrac{1}{2}\\x=\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\) \(\in\) {- \(\dfrac{3}{4};-\dfrac{1}{4}\)}
\(x\) : (- \(\dfrac{1}{3}\))3 = - \(\dfrac{1}{3}\)
\(x\) = (-\(\dfrac{1}{3}\)).(-\(\dfrac{1}{3}\))3
\(x\) = \(\dfrac{1}{81}\)
Vậy \(x=\dfrac{1}{81}\)