Làm ơn viết cái đề rõ hơn dc ko vậy?
\(2014a^2+b^2+c^2\) / \(a^2\) = \(a^2+2014b^2+c^2\) /b\(^2\) = \(a^2+b^2+2014c^2\) /c\(^2\)
P = \(2015a^2+b^2\) /c\(^2\) + \(2015b^2\) +\(c^2\) / a\(^2\) + 2015\(c^2+a^2\)/b\(^2\)
\(\dfrac{2014a^2+b^2+c^2}{a^2}=\dfrac{a^2+2014b^2+c^2}{b^2}=\dfrac{a^2+b^2+2014c^2}{c^2}\)
\(\Rightarrow2014+\dfrac{b^2+c^2}{a^2}=2014+\dfrac{a^2+c^2}{b^2}=2014+\dfrac{a^2+b^2}{c^2}\)
( Lượt 2014 đi )
\(\Rightarrow\dfrac{b^2+c^2}{a^2}=\dfrac{a^2+c^2}{b^2}=\dfrac{a^2+b^2}{c^2}\)
\(\Rightarrow\) \(=\dfrac{b^2+c^2+a^2+c^2+a^2+b^2}{a^2+b^2+c^2}=\dfrac{2.\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=2\)
Làm tiếp bài đầu.
\(\Rightarrow\dfrac{b^2}{a^2}+\dfrac{c^2}{a^2}=\dfrac{a^2}{b^2}+\dfrac{c^2}{b^2}=\dfrac{a^2}{c^2}+\dfrac{b^2}{c^2}=2\)
\(\Rightarrow\dfrac{b^2}{a^2}+\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}+\dfrac{c^2}{b^2}+\dfrac{a^2}{c^2}+\dfrac{b^2}{c^2}=2+2+2=6\)
\(\Rightarrow\dfrac{b^2}{a^2}+\dfrac{c^2}{a^2}+\dfrac{c^2}{b^2}=6:2=3\)
Bài thứ 2.
\(P=\dfrac{2015a^2+b^2}{c^2}+\dfrac{2015b^2+c^2}{a^2}+\dfrac{2015c^2+a^2}{b^2}\)
\(\Rightarrow=2015.\left(\dfrac{a^2}{c^2}+\dfrac{b^2}{a^2}+\dfrac{c^2}{b^2}\right)+\left(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\right)\)
\(=\left(\dfrac{a^2}{c^2}+\dfrac{b^2}{a^2} +\dfrac{c^2}{b^2}\right).\left(2015+1\right)\)
\(=\left(\dfrac{a^2}{c^2}+\dfrac{b^2}{a^2}+\dfrac{c^2}{b^2}\right).2016\)
\(=2016.3=6048\)
Vậy \(P=6048\)
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