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22 tháng 12 2020

\(\left(x-\dfrac{2}{3}\right).\left(x+\dfrac{1}{4}\right)=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\) hoặc \(x+\dfrac{1}{4}=0\)

*) \(x-\dfrac{2}{3}=0\)

\(x=\dfrac{2}{3}\)

*) \(x+\dfrac{1}{4}=0\)

\(x=-\dfrac{1}{4}\)

Vậy \(x=\dfrac{2}{3}\); \(x=-\dfrac{1}{4}\)

=>0,5x=-1-3/4+1+1/2=-1/4

=>x=-0,5

8 tháng 6 2023

`1,5 x -1 1/2 = x -1 3/4`

`=>3/2 x- 3/2 = x- 7/4`

`=> 3/2 x - x = -7/4 +3/2`

`=> (3/2-1)x= -7/4 + 6/4`

`=> ( 3/2 -2/2) x= -1/4`

`=> 1/2 x=-1/4`

`=> x=-1/4 :1/2`

`=> x=-1/4 xx 2`

`=> x= -2/4=-1/2`

 

30 tháng 1 2023

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30 tháng 1 2023

\(x:\dfrac{-31}{15}=\dfrac{-3}{4}-\dfrac{7}{2}\)

\(x:\dfrac{-31}{15}=\dfrac{-3}{4}-\dfrac{14}{4}\)

\(x:\dfrac{-31}{15}=\dfrac{-17}{4}\)

\(x=\dfrac{-17}{4}\times\dfrac{-31}{15}\)

\(x=\dfrac{527}{60}\)

29 tháng 10 2021

3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

13 tháng 10 2021

a) \(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2=\dfrac{4}{25}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)

b) \(\Rightarrow\left(1-\dfrac{1}{4}x\right)^2=\dfrac{121}{49}\)

\(\Rightarrow\left[{}\begin{matrix}1-\dfrac{1}{4}x=\dfrac{11}{7}\\1-\dfrac{1}{4}x=-\dfrac{11}{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{16}{7}\\x=\dfrac{72}{7}\end{matrix}\right.\)

16 tháng 12 2022

a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)

b: =>x-1/2=1/3

=>x=5/6

c: =>2/3x-1=0 hoặc 3/4x+1/2=0

=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3

d =>4/9:x=10/3:9/4=10/3*4/9=40/27

=>x=4/9:40/27=4/9*27/40=108/360=3/10

3 tháng 9 2021

a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)

b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)

c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)

\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)

d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)

\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)

\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)

b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)

\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)

hay x=1

18 tháng 9 2021

1) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{3}{20}\)

2) \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

3) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{4x}=-\dfrac{7}{20}\)

\(\Leftrightarrow4x=-\dfrac{20}{7}\)

\(\Leftrightarrow x=-\dfrac{5}{7}\)

a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|2x-1\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)

25 tháng 7 2017

help me

25 tháng 7 2017

\(\left(x-1\right)\left(x+5\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\end{matrix}\right.\)

\(\left(x-1\right)\left(x+5\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-5< x< 1\)

câu dễ tự làm

\(\Rightarrow x>-5;x< -5\)