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8 tháng 7 2021

( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 1 + 1/3 - 4/3)

=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 4/3 - 4/3)

=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x 0

=0

 

Ta có: \(\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

\(=\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(\dfrac{3}{3}+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

=0

16 tháng 9 2021

\(\dfrac{x-1}{2012}+\dfrac{x-2}{2013}+\dfrac{x-3}{2014}=\dfrac{x-4}{2015}+\dfrac{x-5}{2016}+\dfrac{x-6}{2017}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2012}+1\right)+\left(\dfrac{x-2}{2013}+1\right)+\left(\dfrac{x-3}{2014}+1\right)=\left(\dfrac{x-4}{2015}+1\right)+\left(\dfrac{x-5}{2016}+1\right)+\left(\dfrac{x-6}{2017}+1\right)\)

\(\Leftrightarrow\dfrac{x+2011}{2012}+\dfrac{x+2011}{2013}+\dfrac{x+2011}{2014}-\dfrac{x+2011}{2015}-\dfrac{x+2011}{2016}-\dfrac{x+2011}{2017}=0\)

\(\Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)

\(\Leftrightarrow x=-2011\)( do \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\))

20 tháng 6 2017

Ta có : \(\frac{x+5}{2012}+\frac{x+4}{2013}+\frac{x+3}{2014}=\frac{x+2}{2015}+\frac{x+1}{2016}+\frac{x}{2017}\)

\(\Rightarrow\frac{x+5}{2012}+1+\frac{x+4}{2013}+1+\frac{x+3}{2014}=\frac{x+2}{2015}+1+\frac{x+1}{2016}+1+\frac{x}{2017}+1\)

\(\Leftrightarrow\frac{x+2017}{2012}+\frac{x+2017}{2013}+\frac{x+2017}{2014}=\frac{x+2017}{2015}+\frac{x+2017}{2016}+\frac{x+2017}{2017}\)

\(\Leftrightarrow\frac{x+2017}{2012}+\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}-\frac{x+2017}{2017}=0\)

\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
\(\text{Mà }\)\(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)\ne0\)

\(\text{Nên : }\) x + 2017 = 0 

=> x = -2017

23 tháng 1 2021

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

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

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

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

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

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

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

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

12 tháng 4 2016

cong 1 vao tung hang tu (vd 3/2013) roi dc tong 2016

10 tháng 4 2016

\(\frac{x+3}{2013}+1+\)\(\frac{x+4}{2012}+1+\frac{x+5}{2011}+1\)=\(\frac{x+1}{2015}+1+\frac{x+2}{2014}+1+\frac{x}{2016}+1\)

 \(\Rightarrow\frac{x+2016}{2013}+\frac{x+2016}{2012}+\frac{x+2016}{2011}=\frac{x+2016}{2014}+\frac{x+2016}{2016}\)

\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2016}=0\right)\)

Vì 1/2016+...+1/2011>0 nên (x+2016)=0

suy ra x= -2016

nếu đúng xin kết bạn

ch

11 tháng 4 2016

toán lớp 1 khó thế

11 tháng 5 2023

\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)

\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)

\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)

\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)

\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)

\(\Leftrightarrow x-12=0\)

\(\Leftrightarrow x=12\)