Question

Cho 3 số a,b,c có tổng khác 0 thỏa mãn: a(a^2_bc)+b(b²-ca0+c(c²-ab)=0

tính giá trị biểu thức 

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

Answer 1

\(a\left(a^2-bc\right)+b\left(b^2-ca\right)+c\left(c^2-ab\right)=0\)

\(\Rightarrow a^3-abc+b^3-abc+c^3-abc=0\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\) 

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

Mà \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}\Rightarrow}a=b=c\)

Vậy \(P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)