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(5/2-x)(3x+1)=0

=>\(\left[{}\begin{matrix}\dfrac{5}{2}-x=0\\3x+1=0\end{matrix}\right.\)

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

1 tháng 5 2017

1) (3x-1)(-1/2x+5)=0

TH1: 3x-1=0

        3x   = 1

          x    = 1/3

TH2:  -1/2x+5=0

         -1/2x     =-5

               x     = 10

2)  (3/4-x)^3=-8

     (3/4-x)^3=(-2)^3

=> 3/4-x=-2

           x=3/4+2

           x= 11/4

3)  |2x-1|=-4^2

     |2x-1|=16

=> 2x-1=-16 hoặc 2x-1=16

TH1: 2x-1=-16

        2x   =-15

          x    = -15/2

TH2:  2x-1=16

          2x   =17

            x    = 17/2

19 tháng 6 2023

\(\left(x+2\right)-2=0\)

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

AH
Akai Haruma
Giáo viên
8 tháng 5 2023

Lời giải:
PT $\Leftrightarrow (\frac{x+1}{2022}+1)+(\frac{x+2}{2021}+1)+...+(\frac{x+23}{2000}+1)=0$

$\Leftrightarrow \frac{x+2023}{2022}+\frac{x+2023}{2021}+...+\frac{x+2023}{2000}=0$

$\Leftrightarrow (x+2023)(\frac{1}{2022}+\frac{1}{2021}+...+\frac{1}{2000})=0$
Dễ thấy tổng trong () luôn dương 

$\Rightarrow x+2023=0$

$\Leftrightarrow x=-2023$

30 tháng 9 2016

a) \(\left(2x-1\right)^2-25=0\)

\(\left(2x-1\right)^2=0+25=25\)

\(\left(2x-1\right)^2=5^2=\left(-5\right)^2\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}2x=6\\2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)

b) \(8x^3-50x=0\)

\(2x\left(4x^2-25\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x=0\\4x^2-25=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-\frac{5}{2}\end{array}\right.\end{array}\right.\)

 

3 tháng 5 2021

a) 1/2.x+3/5.(x-2)=3

<=>1/2.x+3/5.x-6/5=3

<=>11/10.x-6/5=3

<=>11/10x=41/10

<=>x=41/11

b) (3x-4).(5x+15)=0

<=>3x-4=0 hoặc 5x+15=0

<=>x=4/3 hoặc x=-3

19 tháng 1 2020

(x-1)2=0

x-1=0

    x=0+1

    x=1

vậy x=1

1 tháng 7 2017

Ko cần đâu bn à mk mong bn đấy

a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)

    \(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)

    \(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)

1 tháng 7 2017

a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0      hay      \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x     = 1         I\(\Leftrightarrow\)\(\frac{-1}{2}\)x     = -5
\(\Leftrightarrow\)  x     = \(\frac{1}{3}\)  I\(\Leftrightarrow\)            x     = 10

b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\)    I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)\(\frac{7}{8}\)          hay     \(\frac{1}{2}x-\frac{1}{3}\)\(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{29}{24}\)        I\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{-13}{24}\)
\(\Leftrightarrow\)      x              = \(\frac{29}{12}\)        I\(\Leftrightarrow\)      x              = \(\frac{-13}{12}\)

c) (2x +\(\frac{3}{5}\))2 - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))2       = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\)         = \(\frac{3}{5}\)    hay      2x +\(\frac{3}{5}\)\(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x                    = 0           I \(\Leftrightarrow\)2x           = \(\frac{-6}{5}\)
\(\Leftrightarrow\)   x                    = 0           I \(\Leftrightarrow\) x           = \(\frac{-3}{5}\)

d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)\(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)

8 tháng 4 2020

(4x-12)(x3+64)=0

=> [x3+64=0=>x=4x-12=0=>4x=12=>x=3           olm bị lỗi nên em đừng có viết cách ra 1 quãng như kia nhé !

vậy x thuộc {3;4}

(3x-12)(x2-4)=0

=>[x2-4=0=>x2=4=>x=2 hoặc x=-23x-12=0=>3x=12=>x=4

vậy x thuộc {4;2;-2}

(x+3)3:3-1=-10

(x+3)3:3=-9

(x+3)3=-9.3

=>(x+3)3=-27

=>x+3=-3

=>x=-6

(3x-1)3-2=-66

(3x-1)3=-64

(3x-1)3=-43

=>3x-1=-4

=>3x=-3

=>x=-1

8 tháng 4 2020

\(\left(4x-12\right)\left(x^3+64\right)=0\)

\(\Leftrightarrow4x-12=0\)

\(\Leftrightarrow4x=0+12\)

\(\Leftrightarrow4x=12\)

\(\Leftrightarrow x=12\div4\)

\(\Leftrightarrow x=3\)

\(\Leftrightarrow x^3+64=0\)

\(\Leftrightarrow x^3=0=64\)

\(\Leftrightarrow x^3=\left(-64\right)\)

\(\Leftrightarrow x^3=\left(-4\right)^3\)

\(\Leftrightarrow x=\left(-4\right)\)

\(\Rightarrow x\in\left\{-4;3\right\}\)

\(\Leftrightarrow\left(3x-12\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow3x-12=0\)

\(\Leftrightarrow3x=0+12\)

\(\Leftrightarrow3x=12\)

\(\Leftrightarrow x=12\div3\)

\(x=4\)

\(\Leftrightarrow x^2-4=0\)

\(\Leftrightarrow x^2=0+4\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x^2=2^2=\left(-2\right)^2\)

\(\Rightarrow x\in\left\{2;-2\right\}\)

\(\Rightarrow x\in\left\{-2;2;4\right\}\)

Các câu khác tương tự nhé !