Question

Найдите значение выражения: (a+2b)²/2(a-4b) + (a-2b)²/3(4b-a) - 6ab+b²/2(a-4b) при a=-7/6,b=1/6

Answer 1

$\frac{(a+2b) ^{2} }{2(a-4b)}+ \frac{(a-2b) ^{2} }{3(4b-a)}- \frac{6ab+b ^{2} }{2(a-4b)}= \\ = \frac{(a+2b) ^{2} }{2(a-4b)}- \frac{(a-2b) ^{2} }{3(a-4b)}- \frac{6ab+b ^{2} }{2(a-4b)}= \\ = \frac{3(a+2b) ^{2}-2(a-2b) ^{2}-3(6ab+b ^{2})}{6(a-4b)}= \\ =\frac{3a ^{2}+12ab+4b ^{2} -2a^{2}+4ab-4b ^{2}-18ab-3b ^{2})}{6(a-4b)}=$$=\frac{a ^{2}-2ab-3b ^{2}}{6(a-4b)} [/tex<span]\\при a=-7/6, b=1/6 получаем [tex] \frac{(- \frac{7}{6}) ^{2}-2\cdot(- \frac{7}{6})( \frac{1}{6})-3\cdot( \frac{1}{6}) {2}}{6(- \frac{7}{6}-4\cdot \frac{1}{6} ) }= \frac{ \frac{49}{46}+ \frac{14}{36}- \frac{3}{36}}{6\cdot(- \frac{11}{6}) } = \frac{ \frac{60}{36} }{-11} =- \frac{10}{66}=- \frac{5}{33}$