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2 tháng 11 2022

\(\left(\dfrac{2}{5}x-2\right):\left(\dfrac{-3}{2}\right)=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{2}{5}x-2=\dfrac{1}{3}.\dfrac{-3}{2}\\ \Leftrightarrow\dfrac{2}{5}x-2=\dfrac{-1}{2}\\ \Leftrightarrow\dfrac{2}{5}x=\dfrac{-1}{2}+2=\dfrac{-3}{2}\\ \Rightarrow x=\dfrac{-3}{2}:\dfrac{2}{5}=\dfrac{-3}{2}.\dfrac{2}{5}=\dfrac{-3}{5}\)

2 tháng 11 2022

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

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

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

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

\(\dfrac{2}{5}x\) = \(\dfrac{3}{2}\)

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

\(x\) = \(\dfrac{15}{4}\)

 

 

29 tháng 10 2021

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

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

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

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

28 tháng 10 2023

a, \(x^2\)  - 19 = 5.9

     \(x^2\) - 19 = 45

     \(x^2\)         = 45 + 19

     \(x^2\)         = 64

      \(x^2\)        = 82

      \(x\)         = 8 

28 tháng 10 2023

b, (2\(x\) + 1)3 = -0,001

    (2\(x\) + 1)3 = (-0,1)3

     2\(x\) + 1   = -0,1

     2\(x\)        = -0,1 - 1

     2\(x\)       = - 1,1

       \(x\)      = -1,1: 2

       \(x\)      = -  0,55

1 tháng 10 2023

\(\left(3-x\right)^3=-\dfrac{27}{64}\)

\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)

\(=>3-x=\dfrac{-3}{4}\)

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

\(x=\dfrac{15}{4}\)

________

\(\left(x-5\right)^3=\dfrac{1}{-27}\)

\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)

\(=>x-5=\dfrac{-1}{3}\)

\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)

\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)

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

\(x=2\)

________

\(\left(2x-1\right)^2=\dfrac{1}{4}\)            

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           hoặc              \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)

\(=>2x-1=\dfrac{1}{2}\)                                       \(2x-1=\dfrac{-1}{2}\)

\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\)                               \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)

\(2x=\dfrac{3}{2}\)                                                     \(2x=\dfrac{1}{2}\)

\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\)                                     \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)

\(x=\dfrac{3}{4}\)                                                       \(x=\dfrac{1}{4}\)

____________

\(\left(2-3x\right)^2=\dfrac{9}{4}\)

\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\)                hoặc                  \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)

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

\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\)                                      \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)

\(3x=\dfrac{1}{2}\)                                                            \(3x=\dfrac{7}{2}\)

\(x=\dfrac{1}{2}.\dfrac{1}{3}\)                                                          \(x=\dfrac{7}{2}.\dfrac{1}{3}\)

\(x=\dfrac{1}{6}\)                                                               \(x=\dfrac{7}{6}\)

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

1 tháng 10 2023

(3-x)3=(-\(\dfrac{3}{4}\))3

3-x=-\(\dfrac{3}{4}\)

  x=3-(-\(\dfrac{3}{4}\))

  x=\(\dfrac{15}{4}\)

11 tháng 9 2017

a) \(\dfrac{2}{5}\)-\(\left(\dfrac{1}{10}-x\right)\)=\(\left(\dfrac{-2}{5}-\dfrac{1}{2}\right)^2\)

\(\dfrac{2}{5}\)- \(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{1}{20}\)

\(\left(\dfrac{1}{10}-x\right)\)= \(\dfrac{2}{5}\)-\(\dfrac{1}{20}\)

\(\left(\dfrac{1}{10}-x\right)\)=\(\dfrac{7}{20}\)

x = \(\dfrac{1}{10}\)-\(\dfrac{7}{20}\)

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

Chúc bn học tốt

30 tháng 10 2023

a) Bổ sung cho đầy đủ đề

b) (3x - 1)/4 = (2x - 5)/3

3(3x - 1) = 4(2x - 5)

9x - 3 = 8x - 20

9x - 8x = -20 + 3

x = -17

c) Điều kiện: x ≠ -1/3

3/(-2) = (x - 3)/(3x + 1)

3.(3x + 1) = -2(x - 3)

9x + 3 = -2x + 6

9x + 2x = 6 - 3

11x = 3

x = 3/11 (nhận)

Vậy x = 3/11

28 tháng 9 2021

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

5 tháng 1 2023

`|x+1/3|+|x+2/3|+|x+2/5|+|x+3/2|=33x`

`@TH1: x >= -1/3`

  `=>x+1/3+x+2/3+x+2/5+x+3/2=33x`

 `=>29x=29/10`

 `=>x=1/10` (t/m)

`@TH2: -2/3 <= x < -1/3`

 `=>-x-1/3+x+2/3+x+2/5+x+3/2=33x`

 `=>31x=67/30`

 `=>x=67/930` (ko t/m)

`@TH3:-2/5 <= x < -2/3`

  `=>-x-1/3-x-2/3+x+2/5+x+3/2=33x`

  `=>33x=9/10`

 `=>x=3/110` (ko t/m)

`@TH4:-3/2 <= x < -2/5`

  `=>-x-1/3-x-2/3-x-2/5+x+3/2=33x`

  `=>35x=1/10`

  `=>x=1/350` (ko t/m)

`@TH5: x < -3/2`

  `=>-x-1/3-x-2/3-x-2/5-x-3/2=33x`

  `=>37x=-29/10`

  `=>x=-29/370` (ko t/m)

5 tháng 1 2023

có VT \(\ge\) 0 với mọi x

=>VP:33x\(\ge\) 0 \(\Rightarrow\) x\(\ge\)0

\(\Rightarrow\) |x+1/3|\(\ge\)0;|x+2/3|\(\ge\) 0;|x+2/5|\(\ge\) 0;|x+3/2|\(\ge\) 0

=> (x+1/3)+(x+2/3)+(x+2/5)+(x+3/2)=33x

=>(x+x+x+x)+(1/3+2/3+2/5+3/2)=33x

=>4x+29/10=33x

=>  29/10=33x-4x

=>29/10=29x

=>x=29/10:29

=>x=1/10

  

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

=>\(\dfrac{1}{4}:x=-\dfrac{5}{2}-\dfrac{3}{4}=-\dfrac{10}{4}-\dfrac{3}{4}=-\dfrac{13}{4}\)

=>\(x=\dfrac{-1}{4}:\dfrac{13}{4}=\dfrac{-1}{4}\cdot\dfrac{4}{13}=\dfrac{-1}{13}\)

b: \(\left(\dfrac{2}{3}\right)^{100}:x=\left(-\dfrac{2}{3}\right)^{98}\)

=>\(\left(\dfrac{2}{3}\right)^{100}:x=\left(\dfrac{2}{3}\right)^{98}\)

=>\(x=\left(\dfrac{2}{3}\right)^{100}:\left(\dfrac{2}{3}\right)^{98}=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)

c: \(\dfrac{3}{2}:\left|4x-\dfrac{1}{5}\right|=\dfrac{3}{4}\)

=>\(\left|4x-\dfrac{1}{5}\right|=\dfrac{3}{2}:\dfrac{3}{4}=\dfrac{3}{2}\cdot\dfrac{4}{3}=2\)

=>\(\left[{}\begin{matrix}4x-\dfrac{1}{5}=2\\4x-\dfrac{1}{5}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{11}{5}\\4x=-\dfrac{9}{5}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{11}{20}\\x=-\dfrac{9}{20}\end{matrix}\right.\)

26 tháng 9 2021

\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

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