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a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 )
<=> 2x - 3 - x + 5 = x + 7 - x - 2
<=> x = 3
b)(7x-5)-(6x+4)=(2x+3)-(2x+1)
<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1
<=> x = 11
c)(9x-3)-(8x+5)=(3x+2)
<=> 9x - 3 - 8x - 5 = 3x + 2
<=> -2x = 10
<=> x = -5
d)(x+7)-(2x+3)=(3x+5)-(2x+4)
<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4
<=> -2x = -3
<=> x = 3/2
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a) \(\left(2x-1\right)+\frac{3}{15}=\frac{3}{2}\)
\(\Rightarrow2x-1=\frac{3}{2}-\frac{3}{15}=\frac{13}{10}\)
\(\Rightarrow2x=\frac{13}{10}+1=\frac{23}{10}\)
\(\Rightarrow x=\frac{23}{20}\)
b) \(x+\frac{46}{15}=1,5\)
\(\Rightarrow x+\frac{46}{15}=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}-\frac{46}{15}\)
\(\Rightarrow x=\frac{-47}{30}\)
c) \(\left(-2x+1\right)+\frac{3}{15}=\frac{5}{3}\)
\(\Rightarrow-2x+1=\frac{5}{3}-\frac{3}{15}=\frac{22}{15}\)
\(\Rightarrow-2x=\frac{7}{15}\Rightarrow x=\frac{-7}{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3x - 7 = 2x + 5
3x - 2x = 5 + 7
x = 12
|3x - 2| = 7
\(\Rightarrow\left\{\begin{matrix}3x-2=7\\-\left(3x-2\right)=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}3x=9\\-3x+2=7\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=3\\-3x=5\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=3\\x=-\frac{5}{3}\end{matrix}\right.\)
4 - |x - 2| = -3
|x - 2| = 4 - (-3)
|x - 2| = 7
\(\Rightarrow\left\{\begin{matrix}x-2=7\\-\left(x-2\right)=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=9\\-x+2=7\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=9\\-x=5\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=9\\x=-5\end{matrix}\right.\)
|2x - 3| = x - 1
\(\Rightarrow\left\{\begin{matrix}2x-3=x-1\\-\left(2x-3\right)=x-1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}2x-x=-1+3\\-2x+3=x-1\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=2\\-2x-x=-1-3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2\\-3x=-4\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=2\\x=\frac{4}{3}\end{matrix}\right.\)
|3x + 1| = x + 3
\(\Rightarrow\left\{\begin{matrix}3x+1=x+3\\-\left(3x+1\right)=x+3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}3x-x=3-1\\-3x-1=x+3\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}2x=2\\-3x-x=3+1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\-4x=4\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\frac{-2}{3}x=\frac{4}{15}\)
x = \(\frac{4}{15}:\frac{-2}{3}\)
x = \(\frac{-2}{5}\)
b, x + \(\frac{1}{3}\) = \(\frac{2}{5}-\left(\frac{-1}{3}\right)\)
x = \(\frac{11}{15}-\frac{1}{3}\)
x = \(\frac{2}{5}\)
c, \(\frac{-2}{3}\)x + \(\frac{-3}{7}\) + \(\frac{1}{2}\)x = \(\frac{-5}{6}\)
\(\frac{-1}{6}\)x + \(\frac{-3}{7}\) = \(\frac{-5}{6}\)
\(\frac{-1}{6}\)x = \(\frac{-5}{6}+\frac{3}{7}\)
\(\frac{-1}{6}\)x = \(\frac{-17}{42}\)
x = \(\frac{-17}{42}:\frac{-1}{6}\)
x = \(\frac{17}{7}\)
d, 2x(x - \(\frac{1}{7}\)) = 0
\(\Rightarrow\) 2x = 0 hoặc x - \(\frac{1}{7}\) = 0
TH1: 2x = 0
x = 0
TH2: x - \(\frac{1}{7}\) = 0
x = \(\frac{1}{7}\)
Vậy x = 0; x = \(\frac{1}{7}\)
Chúc bn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
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 !
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a: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
nên \(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{6}{10}=\dfrac{3}{5}\)
hay \(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b: \(\Leftrightarrow5-\dfrac{4}{7}x=13\)
nên 4/7x=-8
hay x=-12
c: \(\left(x+\dfrac{1}{2}\right)\cdot\left(\dfrac{2}{3}-2x\right)=0\)
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
d: \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
nên 1/6x=5/12
hay x=5/2