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B1 : x + (x+1) + (x+2) + ...+ (x+35) = 0
x + x +1 + x+ 2+...+ x +35 = 0
x + x.35 + (1+2+...+35) = 0
x.36 + 630 =0
x.36 = -630
x = -630 : 36
x =- 17.5
a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)
\(x+\frac{2}{5}=\frac{-7}{20}\)
\(x=\frac{-13}{20}\)
Vậy \(x=\frac{-13}{20}\)
b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)
\(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x=\frac{15}{8}\)
\(x=\frac{-15}{4}\)
Vậy \(x=\frac{-15}{4}\)
\(\frac{-2}{3}\) \(-\) \(\frac{1}{3}\) X \(\left(2.x-5\right)\) \(=\frac{3}{2}\)
\(-1\) X \(\left(2.x-5\right)\) \(=\frac{3}{2}\)
\(\left(2.x-5\right)\) \(=\frac{3}{2}\) \(:-1\)
\(\left(2.x-5\right)\) \(=\frac{3}{2}\)
\(2.x\) \(=\frac{3}{2}\) \(+\) \(5\)
\(2.x\) \(=\frac{7}{2}\)
\(x=\) \(\frac{7}{2}\) \(:2\)
\(x=\frac{7}{4}\)
* Mới lớp 5 nên không chắc, sai thongcam *
#Ninh Nguyễn
\(\frac{-2}{3}-\frac{1}{3}\cdot\left(2x-5\right)=\frac{3}{2}\)
\(\frac{1}{3}\left(2x-5\right)=\frac{-2}{3}-\frac{3}{2}\)
\(2x-5=\frac{-13}{6}:\frac{1}{3}\)
\(2x=\frac{-13}{2}+5\)
\(x=\frac{-3}{2}:2\)
\(x=\frac{-3}{4}\)
1: =>|3x-3/2|=x-1/2+1/4=x-1/4
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{4}\\\left(3x-\dfrac{3}{2}-x+\dfrac{1}{4}\right)\left(3x-\dfrac{3}{2}+x-\dfrac{1}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{4}\\\left(2x-\dfrac{5}{4}\right)\left(4x-\dfrac{7}{4}\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{5}{8};\dfrac{7}{16}\right\}\)
2: =>|1/3x+2/3|=5/3-1+x=x+2/3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{2}{3}\\\left(\dfrac{1}{3}x+\dfrac{2}{3}-x-\dfrac{2}{3}\right)\left(\dfrac{1}{3}x+\dfrac{2}{3}+x+\dfrac{2}{3}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{2}{3}\\\left(-\dfrac{2}{3}x\right)\cdot\left(\dfrac{4}{3}x+\dfrac{4}{3}\right)=0\end{matrix}\right.\Leftrightarrow x=0\)