Question

39 за найдите значение выражения при всех допустимых значениях переменных 2a-b a-2b 6ab-24a^2 - + 3a-b 3a+b b^2-9a^2

Answer 1

$\frac{2a-b}{3a-b}- \frac{2-2b}{3a+b} + \frac{6ab-24a^2}{b^2-9a^2} =\frac{6a^2+2ab-3ab-b^2-3a^2+ab+6ab-2b^2-6ab+24a^2}{(3a-b)(3a+b)} =$$\frac{27a^2-3b^2}{(3a-b)(3a+b)} = \frac{3(9a^2-b^2)}{9a^2-b^2} =3$

Answer 2

$\frac{2a-b}{3a-b}- \frac{a-2b}{3a+b} + \frac{6ab-24a^{2}}{b^{2}-9a ^{2} } = \frac{2a-b}{3a-b}- \frac{a-2b}{3a+b} - \frac{6ab-24a^{2}}{9a ^{2}-b^{2} } = \\ \frac{(2a-b)(3a+b)-(a-2b)(3a-b)-(6ab-24a^{2})}{9a^{2}-b^{2}} = \\ \frac{6 a^{2}+2ab-3ab-b^{2}-3a^{2}+ab+6ab-2b^{2}-6ab+24a^{2} }{9a^{2}-b^{2}}= \\ \frac{27a^{2}-3b^{2}}{9a^{2}-b^{2}} = \frac{3(9a^{2}-b^{2})}{9a^{2}-b^{2}} =3 \\$