Biết b#3a,b#-3a và 6a^2-15ab+5b^2=0.
Tính giá trị biểu thức Q=2a-b/3a-b+5b-a/3a+b
$6a^2-15ab+5b^2=0\Leftrightarrow 15ab=6a^2+5b^2$
$\Rightarrow Q=\dfrac{2a-b}{3a-b}+\dfrac{5b-a}{3a+b}=\dfrac{(2a-b)(3a+b)+(5b-a)(3a-b)}{(3a-b)(3a+b)}$
$=\dfrac{6a^2+2ab-3ab-b^2+15ab-5b^2-3a^2+ab}{9a^2-b^2}$
$=\dfrac{3a^2+15ab-6b^2}{9a^2-b^2}=\dfrac{3a^2+6a^2+5b^2-6b^2}{9a^2-b^2}$
$=\dfrac{9a^2-b^2}{9a^2-b^2}=1$