Question

помогите пожалуйста ​

Answer 1

Ответ:

Объяснение:

2.

a) = $\frac{y}{4x}$

б) = $\frac{3x(x+3)}{3x}$ = x+3

в) = $\frac{2(a-3)}{(a-3)^{2} }$ = $\frac{2}{a-3}$

г) = $\frac{(2y-1)(2y+1)}{(1+2y)(1-y+y^{2})\\}$ = $\frac{2y-1}{1-y+y^{2} }$

3.

a) = $\frac{3y-1+5-8y}{y^{2} }$ = $\frac{4-5y}{y^{2} }$

б) = $\frac{3(x^{2} +3)-x^{2} (3-x)}{3x^{3} }$ = $\frac{3x^{2} +9-3x^{2} +x^{3} }{3x^{3} }$ = $\frac{3}{x^{3} }$ +$\frac{1}{3}$

в) = $\frac{3b}{a(a+b)}$ - $\frac{3a}{b(b+a)}$ = $\frac{3b^{2}-3a^{2} }{ab(a+b)}$ = $\frac{3(a+b)(a-b)}{ab(a+b)}$ = $\frac{3a-3b}{ab}$ = $\frac{3}{b}$ - $\frac{3}{a}$

г) = $\frac{3}{x+3}$ + $\frac{3}{x(x-3)}$ +$\frac{2x}{(3-x)(3+x)}$ = $\frac{3x(x-3)+3(x+3)-2x^{2} }{x(x-3)(x+3)}$ = $\frac{3x^{2} -9x+ 3x+9-2x^{2} }{x(x-3)(x+3)}$= $\frac{x^{2}-6x+9 }{x(x-3)(x+3)}$ = $\frac{(x-3)^{2} }{x(x-3)(x+3)}$ = $\frac{x-3}{x^{2} +3x}$