Question

Упростите дробь 2ab²-10b²/a²-10a+25,найдите значение при b≠1, a=8.​

Answer 1

$\displaystyle\bf\\\frac{2ab^{2}-10b^{2} }{a^{2}-10a+25 } =\frac{2b^{2}\cdot(a-5) }{(a-5)^{2} } =\frac{2b^{2} }{a-5} \\\\\\b=1 \ \ , \ \ a=8\\\\\\\frac{2b^{2} }{a-5} =\frac{2\cdot 1^{2} }{8-5} =\frac{2}{3}$