Question

Tính\(A=\frac{2a-b}{3a-b}+\frac{5b-a}{3a+b}\)biết \(10a^2-3b^2+5ab=0\)và \(9a^2-b^2\ne0\)

Answer 1

Theo giả thiết, ta có:

\(10a^2-3b^2+5ab=0\)

nên   \(3\left(10a^2-3b^2+5ab\right)=0\)

\(\Leftrightarrow\)  \(30a^2-9b^2+15ab=0\)

\(\Leftrightarrow\)  \(15ab=-30a^2+9b^2\)

Do đó:  \(A=\frac{2a-b}{3a-b}+\frac{5b-a}{3a+b}=\frac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}=\frac{3a^2+15ab-6b^2}{9a^2-b^2}=\frac{3a^2+\left(-30a^2+9b^2\right)-6b^2}{9a^2-b^2}\)

             \(A=\frac{-27a^2+3b^2}{9a^2-b^2}=\frac{-3\left(9a^2-b^2\right)}{9a^2-b^2}=-3\)  (do  \(9a^2-b^2\ne0\)  )