1 phần 3 nhân x + 2 phần 3 nhân<x - 2> = 45
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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
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 \(x=\dfrac{1}{81}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{11}{75}:\frac{1}{2}=\frac{22}{75}\Leftrightarrow\frac{1}{x+2}=\frac{1}{25}\Leftrightarrow x=23\)
1/3x + 2/3(x-2) = 45
1/3x + 2/3x - 4/3 = 45
x= 45+4/3
x= 139/3
1/3 x + 2/3 (x-2) = 45
1/3x + 2/3x - 4/3 = 45
x = 45 + 4/3
x = 139/3
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