Thay 2015= x+1 , ta có

f(2014)=...........

tự làm nốt nha mk lười lắm sorry

=4050151.001

2012×2013+2011/2014×2013-2015=2012×2013+2011/(2012+2)×2013-2015=2012×2013+2011/2012×2013+2×2015=2012×2013+2011/2012×2013+4026-2015=2012×2013+2011/2012×2013+2011=1

thay x=2014 vào ta có:

f(2014)=2014²⁰¹⁴-2015.2014²⁰¹³+2015.2014²⁰¹²-2015.2014²⁰¹¹+...-2015.2014+2015

=2014²⁰¹⁴-(2014+1)2014²⁰¹³+(2014+1).2014²⁰¹²-(2014+1).2014²⁰¹¹+...-(2014+1).2014+2014+1

=2014²⁰¹⁴-2014²⁰¹⁴-2014²⁰¹³+2014²⁰¹³+2014²⁰¹²-2014²⁰¹²-2014²⁰¹¹+...-2014²-2014+2014+1

=1

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

toán lớp 1 khó thế

rối loạn mắt rồi hu hu

\(A=\left|x-2011\right|+\left|x-2012\right|+\left|x-2013\right|+\left|x-2014\right|+\left|x-2015\right|\)

\(A=\left|x-2011\right|+\left|x-2012\right|+\left|2014-x\right|+\left|2015-x\right|+\left|x-2013\right|\)

Ta có: \(\left\{{}\begin{matrix}\left|x-2011\right|\ge x-2011\\\left|x-2012\right|\ge x-2012\\\left|2014-x\right|\ge2014-x\\\left|2015-x\right|\ge2015-x\end{matrix}\right.\)

\(A\ge x-2011+x-2012+2014-x+2015-x+\left|x-2013\right|\)

\(A\ge6+\left|x-2013\right|\ge6\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x\ge2011\\x\ge2012\\x\le2014\\x\le2015\end{matrix}\right.\) và \(x=2013\)

\(\Rightarrow\left\{{}\begin{matrix}2012\le x\le2014\\x=2013\end{matrix}\right.\Leftrightarrow x=2013\)

Vậy....

để Anhỏ nhất => x=2013 mình nghĩ thế thôi