Question

1. Сократите дробь:

$\tt a) \ \ \cfrac{22p^4q^2}{99p^5q}; \ \ \ \ \ b) \ \ \cfrac{7a}{a^2+5a}; \ \ \ \ \ c) \ \ \cfrac{x^2-y^2}{4x+4y}$


2. Представьте в виде дроби:

$\tt a) \ \ \cfrac{y-20}{4y}+\cfrac{5y-2}{y^2}; \ \ \ \ \ b) \ \ \cfrac{1}{5c-d}-\cfrac{1}{5c+d}; \ \ \ \ \ c) \ \ \cfrac{7}{a+5}-\cfrac{7a-3}{a^2+5a}$

Answer 1

$\displaystyle \tt 1).\\\\a). \ \ \frac{22p^{4}q^{2}}{99p^{5}q}= \frac{11\cdot2q}{11\cdot9p}= \frac{2q}{9p};\\\\\\b). \ \ \frac{7a}{a^{2}+5a}= \frac{7a}{a(a+5)}= \frac{7}{a+5};\\\\\\c). \ \ \frac{x^{2}-y^{2}}{4x+4y}= \frac{(x-y)(x+y)}{4(x+y)}= \frac{x-y}{4};$

$\displaystyle \tt 2).\\\\a). \ \ \frac{y-20}{4y}+ \frac{5y-2}{y^{2}}= \frac{y(y-20)+4(5y-2)}{4y^{2}}= \frac{y^{2}-20y+20y-8}{4y^{2}}=\frac{y^{2}-8}{4y^{2}};\\\\\\b). \ \ \frac{1}{5c-d}- \frac{1}{5c+d}= \frac{5c+d-(5c-d)}{(5c-d)(5c+d)}= \frac{2d}{25c^{2}-d^{2}};\\\\\\c). \ \ \frac{7}{a+5}- \frac{7a-3}{a^{2}+5a}= \frac{7a-7a+3}{a(a+5)}= \frac{3}{a(a+5)};$