khoan cach giua 2 so la :2-1=1

co tat ca so so hang hay x la :(10-1):1+1=10

(x+1)+(x+2)+.....................(x+10)=2015

x*10+[(1+10)*10/2]=2015

x*10+55               =2015

x*10                     =2015-55

x*10                     =1960

        x                  =1960/10

       x                    =196   

10x+55=2015

10x=1960

x=196

tìm x biết:( x+1) + ( x+ 2) + ( x+3)+ ( x+4)+.....+( x+9)+ ( x+10) = 2015

Giải:Ta có:\(\left(x+1\right)+\left(x+2\right)+......+\left(x+10\right)=2015\)

\(\Rightarrow x+1+x+2+.....+x+10=2015\)

\(\Rightarrow10x+\left(1+2+......+10\right)=2015\)

\(\Rightarrow10x+55=2015\Rightarrow10x=2015-55\)

\(\Rightarrow10x=1960\Rightarrow x=196\)

Vậy .....................

/x+2015/+/x+2019/=4

x+2015+x+2019=4

2x+4034=4

2x=-4030

x=-2015

\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)

=> \(\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)

=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)

=> \(\frac{x+2016}{2012}+\frac{x+2016}{2013}-\frac{x+2016}{2014}-\frac{x+2016}{2015}=0\)

=> \(\left(x+2016\right).\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)

Vì \(\frac{1}{2012}>\frac{1}{2014};\frac{1}{2013}>\frac{1}{2015}\)

=> \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)

=> \(x+2016=0\)

=> \(x=-2016\)

a.

\(\left(-12\right).x-4=5^2.\left(-4\right)\)

\(\Leftrightarrow\left(-12\right).x-4=25.\left(-4\right)\)

\(\Leftrightarrow\left(-12\right).x-4=-100\)

\(\Leftrightarrow\left(-12\right).x=4-100\)

\(\Leftrightarrow\left(-12\right).x=-96\)

\(\Leftrightarrow x=\left(-96\right):\left(-12\right)\)

\(\Leftrightarrow x=8\)

b.

\(\left(-2015+184\right)-\left(84-2015\right)+\left(-200\right)\)

\(=\left(-2015+2015\right)+\left(184-84\right)+\left(-200\right)\)

\(=0+100+\left(-200\right)\)

\(=-100\)

a) (-12).x - 4 = 5².(-4)

-12x - 4 = 25.(-4)

-12x - 4 = -100

-12x = -100 + 4

-12x = -96

x = -96 : (-12)

x = 8

x=8:| (dễ vậy mà):>

\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)

\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)

\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)

\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)

\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)

\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)

\(\Leftrightarrow x=2017\)         \(\text{Vậy }x=2017\)

\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)

\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)

\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)

\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)

b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)

\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)

Phương trình tương đương với: 

\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)

c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)

\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)

\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)