Ta có : \(\frac{2014a^2+b^2+c^2}{a^2}=\frac{a^2+2014b^2+c^2}{b^2}=\frac{a^2+b^2+2014c^2}{c^2}\)

\(\Rightarrow\) \(2014+\frac{b^2+c^2}{a^2}=2014+\frac{a^2+c^2}{b^2}=2014+\frac{a^2+b^2}{c^2}\)

\(\Rightarrow\) \(\frac{b^2+c^2}{a^2}=\frac{a^2+c^2}{b^2}=\frac{a^2+b^2}{c^2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{b^2+c^2}{a^2}=\frac{a^2+c^2}{b^2}=\frac{a^2+b^2}{c^2}=\frac{2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=2\) (Vì \(a^2+b^2+c^2\ne0\))

Suy ra: \(\frac{b^2}{a^2}+\frac{c^2}{a^2}=\frac{a^2}{b^2}+\frac{c^2}{b^2}=\frac{a^2}{c^2}+\frac{b^2}{c^2}=2\)

\(\Rightarrow\frac{b^2}{a^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}+\frac{b^2}{c^2}=2+2+2=6\)

\(\Rightarrow\) \(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}=\frac{6}{2}=3\)

Lại có: \(P=\)\(\frac{2015a^2+b^2}{c^2}+\frac{2015a^2+c^2}{b^2}+\frac{2015b^2+c^2}{a^2}\)

\(=2015\left(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}\right)+\left(\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}\right)\)

\(=\left(2015+1\right)\left(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}\right)\)

\(=2016\left(\frac{a^2}{c^2}+\frac{b^2}{a^2}+\frac{c^2}{b^2}\right)\)

\(=2016.3=6048\)

Vậy \(P=6048\)

1.     (a-b)(a+b)=a²-ab+ab-b²=a²-b²

^2.      (a+b+c)²=(a+b)²+2(a+b)c+c²

                   =a²+2ab+b²+2ac+2bc+c²

                         =a²+b²+c²+2ab+2bc+2ac

A = (-2-1) . (3-1)

A = (-3) . 2

A = -6

B = (-2²-2) . (2²-3)

B = (4 -2) . (4 - 3)

B = 2 . 1

B = 2

C= (-1²-1) . (3.-2 -2)

C= (1-1) . (-6-2)

C= 0 . (-4)

C= 0

Giải:

+ Với \(a=-2\) và \(b=3\), ta có:

\(A=\left(-2-1\right)\left(3-1\right)\)

\(\Leftrightarrow A=-3.2=-6\)

Vậy ...

+ Với \(a=-2\) và \(b=2\), ta có:

\(B=\left[\left(-2\right)^2-2\right]\left(2^2-3\right)\)

\(\Leftrightarrow B=\left(4-2\right)\left(4-3\right)\)

\(\Leftrightarrow B=2.1=2\)

Vậy ...

+ Với \(a=-1\) và \(b=-2\), ta có:

\(C=\left[\left(-1\right)^2-1\right]\left[3.\left(-2\right)-2\right]\)

\(\Leftrightarrow C=\left(1-1\right)\left(-6-2\right)\)

\(\Leftrightarrow C=0.\left(-8\right)=0\)

Vậy ...