Question

Biết \(b\text{≠}\pm-3a\) và \(6a^2-15ab+5b^2=0\).Tính giá trị biểu thức :

\(Q=\dfrac{2a-b}{3a-b}+\dfrac{5b-a}{3a+b}\)

Answer 1

Ta có:

\(Q=\dfrac{2a-b}{3a-b}+\dfrac{5b-a}{3a+b}\)

\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)}{\left(3a-b\right)\left(3a+b\right)}+\dfrac{\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}\)

\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}\)

\(Q=\dfrac{3a^2+15ab-6b^2}{9a^2-b^2}\)

Ta lại có:

\(6a^2-15ab+5b^2=0\)

\(\Rightarrow3a^2+15ab-6b^2=9a^2-b^2\left(1\right)\)

Thay (1) vào Q

=> Q = 1