Question

Помогите пожалуйста даю 35 баллов<нужно решить 3 номера

Answer 1

Ответ:

$\displaystyle 1)\ \ \frac{3x^2}{7x}=\frac{x\cdot 3x}{7x}=\frac{3x}{7}\ \ \ ,\ \ \ \frac{8y^3}{15y^4}=\frac{8\cdot y^3}{15y\cdot y^3}=\frac{8}{15y}\\\\\\\frac{4z^2}{-8z^3}=\frac{4z^2}{4z^2\cdot (-2z)}=-\frac{1}{2z}\ \ \ ,\ \ \ \ \frac{2a^5}{a^4}=\frac{2a^4\cdot a}{a^4}=\frac{2a}{1}=2a\\\\\\\frac{-14b^2}{21b^4}=-\frac{7b^2\cdot 2}{7b^2\cdot 3b^2}=-\frac{2}{3b^2}$

$\displaystyle 2)\ \ \frac{3a+3b}{5(a+b)}=\frac{3(a+b)}{5(a+b)}=\frac{3}{5}\ \ ,\ \ \ \frac{7x-14y}{3x-6y}=\frac{7(x-2y)}{3(x-2y)}=\frac{7}{3}\\\\\\\frac{5a-20c}{15ac}=\frac{5(a-4c)}{5\cdot 3ac}=\frac{a-4c}{3ac}\ \ ,\ \ \ \ \frac{x-2b}{x^2-2bx}=\frac{x-2b}{x(x-2b)}=\frac{1}{x}$

$\displaystyle 3)\ \ \frac{5x-10}{x^2-4}=\frac{5(x-2)}{(x-2)(x+2)}=\frac{5}{x+2}\ \ ,\ \ \ \ \frac{a^2-9}{15+5a}=\frac{(a-3)(a+3)}{5(3+a)}=\frac{a-3}{5}\\\\\\\frac{x^2-4x+4}{3x-6}=\frac{(x-2)^2}{3(x-2)}=\frac{x-2}{3}\ \ ,\ \ \ \frac{b^2+6b+9}{b^2-9}=\frac{(b+3)^2}{(b-3)(b+3)}=\frac{b+3}{b-3}$