\(3x+7⋮x-2\)

\(3\left(x-2\right)+13⋮x-2\)

Vì \(3\left(x-2\right)⋮x-2\)

\(13⋮x-2\)

\(\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)

Ta lập bảng 

cảm ơn nha

a,5x+8=2x-7                         

5x+8-2x+7=0

<=>3x+15=0

<=>3x=-15

<=>x=-5

Vậy x=-5

   b,3.(x+2)=2.(x-1)

<=>3x+6=2x-1

<=>3x+6-2x+1=0

<=>x+7=0

<=>x=-7

Vậy x=-7

a,5x+8=2x-7

   5x - 2x = -7 - 8

    3x        = -15

      x        = (-15) : 3

      x        = -5

Vậy x = -5

 b,3.(x+2)=2.(x-1)

    3x + 6 = 2x - 2

    3x - 2x= -2 - 6

       x     = -8

Vậy x = -8

# HOK TỐT #

x = -4 

k cho mk nhé

\(\left|x+9\right|.2=10\)

\(\Leftrightarrow\left|x+9\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)

Vậy \(x\in\left\{-4;-14\right\}\)

2016/2015 lớn hơn k và kết bạn nha

trả lời

2016/2015 > 2017/2016

nhớ k cho mình nha

học tốt

\(\text{( a-b)-(a+b)+(2a-b)-(2a-3b)=0}\)

\(\Leftrightarrow\text{ a-b-a-b+2a-b-2a+3b = 0}\)

\(\Leftrightarrow\text{0=0}\)

\(\Rightarrow\text{ĐPCM}\)

\(\left(a+b-c\right)-\left(a-b+c\right)+\left(b+c-a\right)-\left(a-b-c\right)=2b\)

\(a+b-c-a+b-c+b+c-a-a+b+c=2b\)

\(-2a+4b-2c=2b\)

\(-2a+4b-2c-2b=0\)

\(-2a+2b-2c=0\)

\(đpcm\) 

\(B=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{2014}}\)

\(4B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)

\(3B=1-\frac{1}{4^{2014}}\)

\(B=\frac{1-\frac{1}{4^{2014}}}{3}\)

\(B=\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2013}}+\frac{1}{4^{2014}}\)

\(4B=1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2012}}+\frac{1}{4^{2013}}\)

\(4B-B=\left(1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{2012}}+\frac{1}{4^{2013}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2013}}+\frac{1}{4^{2014}}\right)\)

\(3B=1-\frac{1}{4^{2014}}\)

\(B=\left(1-\frac{1}{4^{2014}}\right):3\)

a) | x + 2| = 0        

=>x+2=0

        x=0-2

        x=-2

Vậy x=-2

b) | x - 5| = |-7|

    |x-5|=7

\(\Rightarrow\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=7+5=12\\x=-12+5=-7\end{cases}}\)

Vậy..............................................

c) | x - 3 | = 7 - ( -2)                         

    | x-3|=9

* x-3=9             *x-3=-9

      x=9+3              x=-9+3

      x=12                x=-6

Vậy........................................                

d) ( 7 - x) - ( 25 + 7 ) = - 25

    (7-x)-32=-25

     7-x=-25+32

     7-x=7

        x=7-7

        x=0

Vậy x=0

e) | x - 3| = |5| + | -7|

     |x-3|=12

* x-3=12                  * x-3=-12

      x=12+3                    x=-12+3

      x= 15                        x=-9

Vậy...............................................                               

g) 4 - ( 7 - x) = x - ( 13 -4) 

    4-7-x=x-9

    3-x=x-9

    3+9=x+x

     12=2x

       x=12:2

       x=6

Vậy x=6

\(a,|x+2|=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

\(b,|x-5|=|-7|\)

\(\Leftrightarrow|x-5|=7\Leftrightarrow\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}\Rightarrow\orbr{\begin{cases}x=12\\x=-2\end{cases}}}\)

\(c,|x-3|=7-\left(-2\right)\)

\(|x-3|=9\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}}\)

\(d,\left(7-x\right)-\left(25+7\right)=-25\)

\(\left(7-x\right)-32=-25\)

\(7-x=7\Leftrightarrow x=0\)

\(e,|x-3|=|5|+|-7|\)

\(|x-3|=12\Leftrightarrow\orbr{\begin{cases}x-3=12\\x-3=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}x=15\\x=-9\end{cases}}}\)

\(g,4-\left(7-x\right)=x-\left(13-4\right)\)

\(4-7+x=x-13+4\)

\(-3+x=x-17\)

\(x-x=-17+3\)

\(-2x=-14\)ko chắc chỗ này vì -x-x=-2x nha bn 

\(x=7\)