Question

1. Найдите значение выражения:

$\tt\cfrac{14b^2-c}{7b}-2b$ при $\tt b=0,5; \ \ c=-14$


2. Упростите выражение:

$\tt\cfrac{5}{x-7}-\cfrac{2}{x}-\cfrac{3x}{x^2-49}+\cfrac{21}{49-x^2}$

Answer 1

$\displaystyle \tt 1). \ \ \frac{14b^{2}-c}{7b}-2b= \frac{14b^{2}-c-14b^{2}}{7b}=- \frac{c}{7b};\\\\\\npu \ \ b=0,5 \ \ \ \ c=-14; \ \ \ \ \ \ \ \ \ \ -\frac{c}{7b}= \frac{14}{7\cdot0,5}= \frac{2}{0,5}=4;$

$\displaystyle \tt 2). \ \ \frac{5}{x-7}- \frac{2}{x}- \frac{3x}{x^{2}-49}+ \frac{21}{49-x^{2}}= \frac{5x-2x+14}{x(x-7)}- \frac{3x}{x^{2}-49}- \frac{21}{x^{2}-49}=\\\\\\{} \ = \frac{3x+14}{x(x-7)}- \frac{3x+21}{(x-7)(x+7)}= \frac{(x+7)(3x+14)-x(3x+21)}{x(x-7)(x+7)}=\\\\\\{} \ = \frac{3x^{2}+21x+14x+98-3x^{2}-21x}{x(x-7)(x+7)}= \frac{14x+98}{x(x-7)(x+7)}= \frac{14(x+7)}{x(x-7)(x+7)}=\\\\\\{} \ \ =\frac{14}{x(x-7)};$