Question

[Đặng Quốc Huy]

Cho các số nguyên a,b,c thỏa mãn: 2014a^2+b^2+c^2/a^2 = a^2+2014b^2+c^2/b^2 = a^2+b^2+2014c^2/c^2

Tính giá trị biểu thức: P= 2015a^2+b^2/c^2 + 2015b^2+c^2/a^2 + 2015c^2+a^2/b^2

Answer 1

[Ťɧε⚡₣lαsɧ]

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 \(P=6048\)