TA có: \(5ab=3b^2-10a^2\)
=>\(10a^2+5ab-3b^2=0\)
=>\(3a^2+15ab-6b^2+7a^2-10ab+3b^2=0\)
=>\(3a^2+15ab-6b^2=-7a^2+10ab-3b^2\)
\(\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)\)
\(=6a^2+2ab-3ab-b^2+15ab-5b^2-3a^2+ab\)
\(=3a^2-6b^2+15ab=3\left(a^2+5ab-2b^2\right)\)
\(10a^2+5ab-3b^2=0\)
=>\(9a^2-b^2+a^2+5ab-2b^2=0\)
=>\(9a^2-b^2=-a^2-5ab+2b^2=-\left(a^2+5ab-2b^2\right)\)
Ta có: \(A=\frac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{9a^2-b^2}\)
\(=\frac{3\cdot\left(a^2+5ab-2b^2\right)}{-\left(a^2+5ab-2b^2\right)}=-3\)
Đúng 0
Bình luận (0)
