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\(\Leftrightarrow\frac{x^{2014}}{a^2+b^2+c^2+d^2}+\frac{y^{2014}}{a^2+b^2+c^2+d^2}+\frac{z^{2014}}{a^2+b^2+c^2+d^2}+\frac{t^{2014}}{a^2+b^2+c^2+d^2}\)
\(-\frac{x^{2014}}{a^2}-\frac{y^{2014}}{b^2}-\frac{z^{2014}}{c^2}-\frac{t^{2014}}{d^2}=0\)
\(\Leftrightarrow\left(\frac{x^{2014}}{a^2+b^2+c^2+d^2}-\frac{x^{2014}}{a^2}\right)+\left(\frac{y^{2014}}{a^2+b^2+c^2+d^2}-\frac{y^{2014}}{b^2}\right)+\left(\frac{z^{2014}}{a^2+b^2+c^2+d^2}-\frac{z^{2014}}{c^2}\right)\)
\(+\left(\frac{t^{2014}}{a^2+b^2+c^2+d^2}-\frac{t^{2014}}{d^2}\right)=0\)
\(\Leftrightarrow x^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\right)+y^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\right)+\)
\(z^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\right)+t^{2014}.\left(\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\right)=0\)
vì a2,b2,c2,d2 lớn hơn hoặc bằng 0
=> \(\hept{\begin{cases}\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{a^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{b^2}\ne0\\\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{c^2}\ne0\end{cases}}và....\frac{1}{a^2+b^2+c^2+d^2}-\frac{1}{d^2}\ne0\)
\(\Rightarrow\hept{\begin{cases}x^{2014}=0\\y^{2014}=0\\z^{2014}=0\end{cases}}và..t^{2014}=0\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}và...t=0\)
=> \(\hept{\begin{cases}x^{2015}=0\\y^{2015}=0\\z^{2015}=0\end{cases}}và..t^{2015}=0\Rightarrow x^{2015}+y^{2015}+z^{2015}+t^{2015}=0\)
vậy \(x^{2015}+y^{2015}+z^{2015}+t^{2015}=0\)
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
Ta có : \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}\)
Mà : \(\left(\frac{2014}{2015}+\frac{1}{2014}\right)>1;\left(\frac{2015}{2016}+\frac{1}{2014}\right)>1;\frac{2014}{2014}=1\)
Nên : \(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}\)\(>1+1+1=3\)
Ta có:\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)\)\(+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}\)
Mà:\(\left(\frac{2014}{2015}+\frac{1}{2014}\right)>1:\left(\frac{2015}{2016}+\frac{1}{2014}\right)>\)\(1:\frac{2014}{2014}=1\)
Nên:\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}=\left(\frac{2014}{2015}+\frac{1}{2014}\right)\)\(+\left(\frac{2015}{2016}+\frac{1}{2014}\right)+\frac{2014}{2014}>1+1+1=3\)
Dễ ợt nhưng éo không biết làm