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20 tháng 3 2017

\(\Leftrightarrow\)3(x-1)-(x+1)=-1

\(\Leftrightarrow\)3x-3-x-1=-1

\(\Leftrightarrow\)3x-x  =  -1+1+3

\(\Leftrightarrow\)2x     =  3

\(\Leftrightarrow\) x      =  \(\frac{3}{2}\)

20 tháng 3 2017

đúng rồi

31 tháng 3 2019

diễn giải ra nhé

14 tháng 5 2022

X x \(\dfrac{3}{4}\)+ X x\(\dfrac{1}{5}\)+ X x \(\dfrac{1}{20}\)+ X= 1000

 

15 tháng 6 2023

\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)

\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{3}{7}\)

\(=\dfrac{6}{35}\)

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

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

\(x=\dfrac{9}{12}-\dfrac{2}{12}\)

\(x=\dfrac{7}{12}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)

\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)

\(\dfrac{1}{2020}+x=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}-\dfrac{1}{2020}\)

\(x=\dfrac{1010}{2020}-\dfrac{1}{2020}\)

\(x=\dfrac{1009}{2020}\)

15 tháng 6 2023

\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)

\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{3}{7}\)

\(=\dfrac{6}{35}\)

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

\(\Rightarrow\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}-x\)

\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1\times2\times3\times4\times...\times2019}{2\times3\times4\times5\times...\times2020}+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2020}+x=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}-\dfrac{1}{2020}=\dfrac{1009}{2020}\)

23 tháng 8 2020

Ta có: \(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+\frac{1}{\left(x+2\right).\left(x+3\right)}-\frac{1}{x}=\frac{1}{2020}\)

    \(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2020}\)

    \(\Leftrightarrow-\frac{1}{\left(x+3\right)}=\frac{1}{2020}\)

    \(\Rightarrow-\left(x+3\right)=2020\)

    \(\Leftrightarrow-x-3=2020\)

    \(\Leftrightarrow-x=2023\)

    \(\Leftrightarrow x=-2023\)

Vậy \(x=-2023\)

23 tháng 8 2020

Bài làm:

Ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2020}\)

\(\Leftrightarrow\frac{\left(x+1\right)-x}{x\left(x+1\right)}+\frac{\left(x+2\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}+\frac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2020}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2020}\)

\(\Rightarrow\frac{1}{-x-3}=\frac{1}{2020}\)

\(\Rightarrow-x-3=2020\Rightarrow x=-2023\)

28 tháng 8 2019

1)1/9 x 3x = 2187:81=27

            3x=27:1/9=243=35

            =>x=5

  

28 tháng 8 2019

\(\frac{1}{9}.3^4.3^x=3^7\)

\(\Leftrightarrow3^x=3^7:\frac{1}{9}:3^4=243\)

\(\Leftrightarrow3^x=3^5\)

\(\Leftrightarrow x=5\)

9 tháng 2 2020

a) \(\frac{8}{9}\cdot x-\frac{2}{3}=\frac{1}{3}\cdot x+1\frac{1}{3}\)

=> \(\frac{8x}{9}-\frac{2}{3}=\frac{x}{3}+\frac{4}{3}\)

=> \(\frac{8x}{9}-\frac{6}{9}=\frac{x+4}{3}\)

=> \(\frac{8x-6}{9}=\frac{x+4}{3}\)

=> \(3\left(8x-6\right)=9\left(x+4\right)\)

=> \(24x-18=9x+36\)

=> \(24x-18-9x=36\)

=> \(24x-9x=54\)

=> \(15x=54\)

=> \(5x=18\)

=> \(x=\frac{18}{5}\)

Vậy x = \(\frac{18}{5}\)

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

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

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=\frac{3}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}:2=\frac{3}{4}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{2};\frac{3}{4}\right\}\)

1: \(x\left(x-1\right)+\left(1+x\right)^2\)

\(=x^2-x+x^2+2x+1\)

\(=2x^2+x+1\)

Đa thức này ko phân tích được nha bạn

2: \(\left(x+1\right)^2-3\left(x+1\right)\)

\(=\left(x+1\right)\cdot\left(x+1\right)-\left(x+1\right)\cdot3\)

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

\(=\left(x+1\right)\left(x-2\right)\)

3: \(2x\cdot\left(x-2\right)-\left(x-2\right)^2\)

\(=2x\left(x-2\right)-\left(x-2\right)\cdot\left(x-2\right)\)

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

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

4: \(3x\left(x-1\right)^2-\left(1-x\right)^3\)

\(=3x\left(x-1\right)^2+\left(x-1\right)^3\)

\(=3x\left(x-1\right)^2+\left(x-1\right)^2\cdot\left(x-1\right)\)

\(=\left(x-1\right)^2\cdot\left(3x+x-1\right)\)

\(=\left(x-1\right)^2\cdot\left(4x-1\right)\)

5: \(3x\left(x+2\right)-5\left(x+2\right)^2\)

\(=\left(x+2\right)\cdot3x-\left(x+2\right)\cdot\left(5x+10\right)\)

\(=\left(x+2\right)\left(3x-5x-10\right)\)

\(=\left(-2x-10\right)\left(x+2\right)\)

\(=-2\left(x+5\right)\left(x+2\right)\)

6: \(4x\left(x-y\right)+3\left(y-x\right)^2\)

\(=4x\left(x-y\right)+3\left(x-y\right)^2\)

\(=\left(x-y\right)\cdot4x+\left(x-y\right)\left(3x-3y\right)\)

\(=\left(x-y\right)\cdot\left(4x+3x-3y\right)\)

\(=\left(x-y\right)\left(7x-3y\right)\)

4 tháng 12 2023

Cảm ơn nhiều

20 tháng 9 2023

:((

26 tháng 9 2016

1 / 5 + x = 3 / 7 + 1 / 3

1 / 5 + x = 16 /21

x            = 16 / 21 - 1 / 5

x            = 59 / 105

x - 1 / 2 = 2 / 3 - 1 / 5

x - 1 / 2 = 7 / 15

         x  = 7 / 15 + 1 / 2

         x  = 29 / 30

3 / 5 * x = 2 / 7+ 1 / 4

3 / 5 * x = 15 / 28

          x = 15 / 28 : 3 / 5

          x = 25 / 28

7 / 8 : x = 1 / 6 * 2 / 3

7 / 8 : x = 1 / 9

          x = 7 / 8 : 1 / 9

          x = 63 / 8