Question

Найдите значение выражения 3×(1/6a-1/7b)/(b/6-a/7) при а=√18, и в=1/√2

Answer 1

$\frac{3*(\frac{1}{6}a-\frac{1}{7}b) }{\frac{b}{6}-\frac{a}{7}} =\frac{3*\frac{7a-6b}{42} }{\frac{7b-6a}{42} }=\frac{3*(7a-6b)}{7b-6a} \\\\a=\sqrt{18}=\sqrt{9*2}=3\sqrt{2} \\\\b=\frac{1}{\sqrt{2} }=\frac{\sqrt{2} }{2} \\\\\\\frac{3*(7a-6b)}{7b-6a}=\frac{3*(7*3\sqrt{2}-6*\frac{\sqrt{2}}{2})}{7*\frac{\sqrt{2} }{2}-6*3\sqrt{2}}=\frac{3*(21\sqrt{2}-3\sqrt{2})}{3,5\sqrt{2}-18\sqrt{2}}=-\frac{3*18\sqrt{2} }{14,5\sqrt{2}} = -\frac{108}{29}=-3\frac{21}{29}$

Если задание выглядит по другому, то вариант решения такой :

$\frac{3*(\frac{1}{6a}-\frac{1}{7b})}{\frac{b}{6}-\frac{a}{7}}=\frac{3*\frac{7b-6a}{6a*7b} }{\frac{7b-6a}{42}} =\frac{3*(7b-6a)*42}{42ab*(7b-6a)}=\frac{3}{a*b}\\\\a=\sqrt{18}=\sqrt{9*2}=3\sqrt{2} \\\\b=\frac{1}{\sqrt{2}}\\\\\\\frac{3}{a*b}=\frac{3}{3\sqrt{2}*\frac{1}{\sqrt{2}}}=\frac{3}{3}=\boxed1$