Từ \(\frac{x-2016}{100}\rightarrow\frac{x-2}{2114}\) có tất cả \(1008\) số.

Ta có: \(\frac{x-2016}{100}+\frac{x-2014}{102}+...+\frac{x-2}{2114}=1008\)

\(\Leftrightarrow\frac{x-2016}{100}+\frac{x-2014}{102}+...+\frac{x-2}{2114}-1008=1008-1008\)

\(\Leftrightarrow\left(\frac{x-2016}{100}-1\right)+\left(\frac{x-2014}{102}-1\right)+...+\left(\frac{x-2}{2114}-1\right)=0\) (*)

\(\Leftrightarrow\frac{x-2116}{100}+\frac{x-2116}{102}+...+\frac{x-2116}{2114}=0\)

\(\Leftrightarrow\left(x-2116\right)\left(\frac{1}{100}+\frac{1}{102}+...+\frac{1}{2114}\right)=0\)

Vì \(\frac{1}{100}+\frac{1}{102}+...+\frac{1}{2114}\ne0\)

Nên từ pt (*) \(\Leftrightarrow x-2116=0\Leftrightarrow x=2116\)

Vậy...

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0.\)

\(1+\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}-4+\frac{x+349}{5}=0\)

\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

\(\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

Study well 

\(\Rightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+349}{5}-4=0\)

\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Rightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Mà \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

Nên \(x+329=0\Rightarrow x=-329\)

Vậy \(x=-329\)

Chúc bạn học tốt !!!

Bài làm

Không viết lại đề nha !

\(=\frac{\left(x+2\right)}{327}+\frac{\left(x+3\right)}{326}+\frac{\left(x+4\right)}{325}+\frac{\left(x+5\right)}{324}+\frac{\left(x+349\right)}{5}=0\)

\(\Leftrightarrow\frac{\left(x+2\right)}{327}+1+\frac{\left(x+3\right)}{326}+1+\frac{\left(x+4\right)}{325}+1+\frac{\left(x+5\right)}{324}+1+\frac{\left(x+349\right)}{5}\)

\(\Leftrightarrow\frac{\left(x+329\right)}{327}+\frac{\left(x+329\right)}{326}+\frac{\left(x+329\right)}{325}+\frac{\left(x+329\right)}{324}+\frac{\left(x+329\right)}{5}-4=0\)

\(\Leftrightarrow\left(x+329\right).\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

Nên:x+329=0=>x=-329

\(\frac{x+2}{327}+\frac{x+3}{326}+....+\frac{x+349}{5}=\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+329}{5}+4=\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\frac{x+329}{5}=\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+....+\frac{1}{324}+\frac{1}{5}\right)=0\Rightarrow x+329=0\Leftrightarrow x=-329\)

Bạn tham khảo tại đây nhé: Câu hỏi của trần như.

Chúc bạn học tốt!

\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)

\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)

Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)

<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)

<=> \(\frac{x-2017}{2016}+\frac{x-2017}{2015}+...+\frac{x-2017}{1}=0\)

<=> \(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}\right)=0\)

<=> \(x-2017=0\)\(\left(do\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}>0\right)\)

<=> \(x=2017\)

Vậy x = 2017

đúng thì

Ta có: \(\hept{\begin{cases}x^2+y^2=1\\\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\end{cases}}\)

\(\Leftrightarrow b\left(a+b\right)x^4+a\left(a+b\right)y^4=ab\left(x^4+2x^2y^2+y^4\right)\)

\(\Leftrightarrow b^2x^4+a^2y^4-2abx^2y^2=0\)

\(\Leftrightarrow\left(bx^2-ay^2\right)^2=0\)

\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow\frac{x^{2016}}{a^{1008}}=\frac{y^{2016}}{b^{1008}}=\frac{1}{\left(a+b\right)^{1008}}\)

\(\Rightarrow\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{21008}}=\frac{2}{\left(a+b\right)^{1008}}\)

Em vào câu hỏi tương tự tham khảo: 

Ta có: \(x^2+y^2=1\Leftrightarrow x^4+2x^2y^2+y^4=1\)

Khi đó: \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{x^4+2x^2y^2+y^4}{a+b}\)

<=> \(\left(a+b\right)\left(\frac{x^4}{a}+\frac{y^4}{b}\right)=x^4+2x^2y^2+y^4\)

<=> \(\frac{b}{a}x^4+\frac{a}{b}y^4=2x^2y^2\)

<=> \(\frac{x^4}{a^2}+\frac{y^4}{b^2}-\frac{2x^2y^2}{ab}=0\)

<=> \(\left(\frac{x^2}{a}-\frac{y^2}{b}\right)^2=0\)

<=> \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)( dãy tỉ số bằng nhau)

Khi đó: \(\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{1008}}=2\frac{x^{2016}}{a^{1008}}=\frac{2}{\left(a+b\right)^{1008}}\)

\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)

\(\Rightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=2016-2016\)

\(\Rightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Rightarrow\left(x-2017\right).\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\right)=0\)

Mà \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+1\ne0\Rightarrow x-2017=0\)

=> x = 2017