Đề ??

\(bx^2=ay^{2^{ }}=\dfrac{x^2}{\dfrac{1}{b}}=\dfrac{y^2}{\dfrac{1}{a}}=\dfrac{x^2+y^2}{\dfrac{a+b}{ab}}=\dfrac{ab}{a+b}.\)

\(\Leftrightarrow\dfrac{x^2}{a}=\dfrac{1}{a+b}=\dfrac{y^2}{b}.\)

\(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\left(\dfrac{x^2}{a}\right)^{1008}+\left(\dfrac{y^2}{b}\right)^{1008}=2.\left(\dfrac{1}{a+b}\right)^{1008}=\dfrac{2}{\left(a +b\right)^{1008}}\left(dpcm\right)\)

Theo bài ra ta có:

\(bx^2=ay^2\) \(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\)

\(x^2+y^2=1\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

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

\(\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{1}{a+b}\) \(\left(1\right)\)

Từ \(\left(1\right)\) suy ra :

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

\(=\left(\dfrac{x^2}{a}\right)^{1008}+\left(\dfrac{y^2}{b}\right)^{1008}\)

\(=\left(\dfrac{1}{a+b}\right)^{1008}+\left(\dfrac{1}{a+b}\right)^{1008}\)

\(=2\cdot\left(\dfrac{1}{a+b}\right)^{1008}\)

\(=2\cdot\dfrac{1^{1008}}{\left(a+b\right)^{1008}}\)

\(=2\cdot\dfrac{1}{\left(a+b\right)^{1008}}\)

\(=\dfrac{2}{a+b}^{1008}\)

Vậy \(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\dfrac{2}{a+b}^{1008}\)

\(\left(x^2+2015x\right)\left(\dfrac{1}{2016}+\dfrac{1}{1008}+\dfrac{1}{672}+1\right)=2022\)

\(\Leftrightarrow\left(x^2+2015x\right).\dfrac{2022}{2016}=2022\)

\(\Leftrightarrow x^2+2015x=2016\)

\(\Leftrightarrow x^2+2015x-2016=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2016\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2016\end{matrix}\right.\)

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

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

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

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

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

\(\Leftrightarrow x-12=0\)

\(\Leftrightarrow x=12\)

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

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

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

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

\(\Leftrightarrow x-2017=0\) (do \(\dfrac{1}{2016}+\dfrac{1}{2015}+...+1\ne0\))

\(\Rightarrow x=2017\)

Phương trình =2016. Mình quên ghi

Giải:

Ta có:

\(\dfrac{x}{2}+\dfrac{x}{4}+\dfrac{x}{2016}=\dfrac{x}{3}+\dfrac{x}{5}+\dfrac{x}{2017}\)

\(\Leftrightarrow\dfrac{x}{2}+\dfrac{x}{4}+\dfrac{x}{2016}-\dfrac{x}{3}-\dfrac{x}{5}-\dfrac{x}{2017}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{2016}-\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{2017}\right)=0\)

Mà \(\dfrac{1}{2}>\dfrac{1}{3};\dfrac{1}{4}>\dfrac{1}{5};\dfrac{1}{2016}>\dfrac{1}{2017}\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{2016}-\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{2017}\right)\) \(\ne0\)

\(\Leftrightarrow x=0\)

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

\(\Leftrightarrow\dfrac{x}{2}+\dfrac{x}{4}+\dfrac{x}{2016}-\dfrac{x}{3}-\dfrac{x}{4}-\dfrac{x}{2017}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{2016}-\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{2017}\right)=0\)

\(\Leftrightarrow x=0\)

Vậy x = 0