Question

Cho a > b > 0, biết \(3a^2+3b^2=10ab\). Tính \(P=\dfrac{a-b}{a+b}\)

Answer 1

Ta có : \(3a^2+3b^2=10ab\)

\(\Leftrightarrow3a^2-ab-9ab+3b^2=0\)

\(\Leftrightarrow a\left(3a-b\right)-3b\left(3a-b\right)=0\)

\(\Leftrightarrow\left(3a-b\right)\left(a-3b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3a-b=0\\a-3b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=3a\left(L\right)\\a=3b\left(N\right)\end{matrix}\right.\)

Thế \(a=3b\) vào P ta được :

\(P=\dfrac{3b-b}{3b+b}=\dfrac{2b}{4b}=\dfrac{1}{2}\)