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1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
a) \(\frac{-2}{3}x+\frac{1}{5}=\frac{1}{10}\)
\(\Leftrightarrow\frac{-2}{3}x=\frac{1}{10}-\frac{1}{5}\)
\(\Leftrightarrow\frac{-2}{3}x=\frac{-1}{10}\)
\(\Leftrightarrow x=\frac{-1}{10}\div\frac{-2}{3}\)
\(\Leftrightarrow x=\frac{3}{20}\)
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !
a. \(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)-\dfrac{3}{2}=\dfrac{1}{4}\\ \Leftrightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \Leftrightarrow\dfrac{1}{2}x=\dfrac{29}{24}\\ \Leftrightarrow x=\dfrac{29}{12}\)
b. \(\dfrac{3}{4}-2\left(2x-\dfrac{2}{3}\right)=2\\ \Leftrightarrow2\left(2x-\dfrac{2}{3}\right)=-\dfrac{5}{4}\\ \Leftrightarrow2x-\dfrac{2}{3}=-\dfrac{5}{8}\\ \Leftrightarrow2x=\dfrac{1}{24}\\ \Leftrightarrow x=\dfrac{1}{48}\)
c. \(\left(2x-3\right)\left(6-2x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
Vậy ..........
1) Ta có: \(3\left(x-1\right)-5\left(x-2\right)=4\left(x+1\right)\)
\(\Leftrightarrow3x-5-5x+10-4x-4=0\)
\(\Leftrightarrow-6x+1=0\)
\(\Leftrightarrow-6x=-1\)
hay \(x=\dfrac{1}{6}\)
2) Ta có: \(-2\left(x-2\right)-4\left(x+1\right)=-3\left(x+3\right)\)
\(\Leftrightarrow-2x+4-4x-4+3x+9=0\)
\(\Leftrightarrow-3x=-9\)
hay x=3
3) Ta có: \(3x^2+2x=0\)
\(\Leftrightarrow x\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
4) Ta có: \(x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) Ta có: \(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
6) Ta có: \(\left(5x-1\right)^3=125\)
\(\Leftrightarrow5x-1=5\)
\(\Leftrightarrow5x=6\)
hay \(x=\dfrac{6}{5}\)
7) Ta có: \(3^{x+1}=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
\(1,\left(\left(3\cdot\left(3x+2\right)\right)+4\cdot\left(x+2\right)\right)-90=0.\)
\(13x-76=0\)
\(13x=76\)
\(x=\frac{76}{13}\)
\(2,\left(\left(5\cdot\left(x+1\right)\right)+3\cdot\left(2x+1\right)\right)-65=0\)
\(11x-57=0\)
\(11x=57\)
\(x=\frac{57}{11}\)
\(3,\left(\left(3\cdot\left(3x+4\right)\right)+5\cdot\left(2x+1\right)\right)-100=0\)
\(19x-83=0\)
\(19x=83\)
\(x=\frac{83}{19}\)
2( \(\dfrac{1}{2}\) x - \(\dfrac{1}{3}\)) - \(\dfrac{3}{2}\) = \(\dfrac{1}{4}\)
2(\(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)) = \(\dfrac{1}{4}\) + \(\dfrac{3}{2}\)
2(\(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)) = \(\dfrac{7}{4}\)
\(\dfrac{1}{2}\)x = \(\dfrac{7}{4}\) : 2 + \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\)x = \(\dfrac{29}{24}\)
x = \(\dfrac{29}{24}\) x2
x = \(\dfrac{29}{12}\)
`2.(1/2 x - 1/3) - 3/2 = 1/4`
`2.(1/2 x - 1/3) = 1/4 + 3/2`
`2.(1/2 x - 1/3) = 1/4 + 6/4`
`2.(1/2 x - 1/3) = 7/4`
`1/2x - 1/3 = 7/4 : 2`
`1/2 x - 1/3 = 7/4 xx 1/2`
`1/2 x - 1/3 = 7/8`
`1/2 x = 7/8 + 1/3`
`1/2 x = 21/28 + 7/28`
`1/2 x = 29/24`
`x=29/24 : 1/2`
`x=29/24 xx 2`
`x=29/12`
Vậy `x = 29/12`