chứng minh rằng a^2016+b^2016<(=)c^2016 với a;b;c là độ dài các cạnh trong tam giác vuông Và c là độ dài cạnh huyền
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\(\frac{a+2015}{a-2015}=\frac{b+2016}{b-2016}\Rightarrow\)\(\frac{a+2015}{a-2015}-1=\frac{b+2016}{b-2016}-1\)
\(\frac{a+2015-a+2015}{a-2015}=\frac{b+2016-b+2016}{b-2016}\Rightarrow\)\(\frac{2015}{a-2015}=\frac{2016}{b-2016}\Rightarrow\)
2015(b-2016) =2016(a-2015) =>2015b =2016a =>\(\frac{a}{b}=\frac{2015}{2016}\)
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ta có \(\frac{a+2015}{a-2015}=\frac{b+2016}{b-2016}\)
\(\Rightarrow\frac{\left(a+2015\right)-\left(a-2015\right)}{a-2015}=\frac{\left(b+2016\right)-\left(b-2016\right)}{b-2016}\)
\(\Rightarrow\frac{2015.2}{a-2015}=\frac{2016.2}{b-2016}\)
\(\Rightarrow\frac{2015}{a-2015}=\frac{2016}{b-2016}\)
\(\Rightarrow\frac{a-2015}{2015}=\frac{b-2016}{2016}\)
\(\Rightarrow\frac{a}{2015}-1=\frac{b}{2016}-1\)
Suy ra ĐPCM
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\(A=\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\\ \Rightarrow A>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\left(1\right)\\ A< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow1< A< B\\ \Rightarrow A\notin Z\)
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\(\Leftrightarrow\left(a^{2016}+b^{2016}\right).\left(c^{2016}-d^{2016}\right)=\left(a^{2016}-b^{2016}\right).\left(c^{2016}+d^{2016}\right)\)
\(\Leftrightarrow ac^{2016}-ad^{2016}+bc^{2016}-bd^{2016}=ac^{2016}+ad^{2016}-bc^{2016}-bd^{2016}\)
\(\Leftrightarrow-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}\)
nếu \(-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}=0\)
\(\Rightarrow ad^{2016}-bc^{2016}=0\Rightarrow ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}\left(1\right)\)
nếu \(\text{}-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}\ne0\Rightarrow ad=-bc\Rightarrow\frac{a}{b}=-\frac{c}{d}\left(2\right)\)
từ (1) và (2) => đpcm