Помогите, пожалуйста, решить! :) № 26 (а, в, д), №27 (а, в,д), №28 (а, в, д) и №36.. Created by 1pc1. algebra-ru

Помогите, пожалуйста, решить! :) № 26 (а, в, д), №27 (а, в,д), №28 (а, в, д) и №...

Akimka $a) \frac{ax-bx}{mx+nx}= \frac{x*(a-b)}{x(m+n)} = \frac{a-b}{m+n}$
$\frac{ x^{2} -2x}{xy-2y} = \frac{x(x-2)}{y(x-2)}= \frac{x}{y}$
$\frac{ a^{2}+ab}{ a^{2}+ ac} = \frac{a(a+b)}{a(a+c)}= \frac{a+b}{a+c}$
№27
$\frac{ y^{3}}{ y^{3}+ y^{2}}= \frac{ y^{3}}{ y^{2}(y+1)}= \frac{y}{y+1}$
$\frac{x^{3} +x^{5}}{x^{2}+1} = \frac{x^{3}(1+ x^{2} )}{ x^{2} +1} = x^{3}$
$\frac{ m^{4}-m^{3}}{m^{2}+m^{3}} = \frac{m^{3}(m-1)}{m^{2}(1+m)} = \frac{m(m-1)}{(1+m)}$
№28
$\frac{ x^{2}-9 y^{2} }{x+3y} = \frac{(x-3y)(x+3y)}{x+3y} =x-3y$
$\frac{m^{2}- n^{2} }{mn- n^{2} } = \frac{(m-n)(m+n)}{n(m-n)} = \frac{m+n}{n}$
$\frac{2ab-6a}{ b^{2}-6b+9 } = \frac{2a(b-3)}{ (b-3)^{2}} = \frac{2a}{b-3}$

Ирэнушка №26(а,в,д)
а) $\frac{ax-bx}{mx+nx}=\frac{x(a-b)}{x(m+n)}=\frac{a-b}{m+n};$

в) $\frac{x^2-2x}{xy-2y}=\frac{x(x-2)}{y(x-2)}=\frac{x}{y};$

д) $\frac{a^2+ab}{a^2+ac}=\frac{a(a+b)}{a(a+c)}=\frac{a+b}{a+c};$

№27(а,в,д)
a) $\frac{y^3}{y^3+y^2}=\frac{y^3}{y^2(y+1)}=\frac{y}{y+1};$

в) $\frac{x^5+x^3}{x^2+1}=\frac{x^3(x^2+1)}{x^2+1}=x^3;$

д) $\frac{m^4-m^3}{m^2+m^3}=\frac{m^3(m-1)}{m^2(1+m)}=\frac{m^3(m-1)}{-m^2(m-1)}=-m;$

№28(а,в,д)
а) $\frac{x^2-9y^2}{x+3y}=\frac{(x-3y)(x+3y)}{x+3y}=x-3y;$

в) $\frac{m^2-n^2}{mn-n^2}=\frac{(m-n)(m+n)}{n(m-n)}=\frac{m+n}{n};$

д) $\frac{2ab-6a}{b^2-6b+9}=\frac{2a(b-3)}{(b-3)^2}=\frac{2a}{b-3};$

№36
а) $\frac{x^3-xy^2}{x^2-xy}=\frac{x(x^2-y^2)}{x(x-y)}=\frac{(x-y)(x+y)}{(x-y)}=x+y;$

б) $\frac{2z^2-8}{6z^2+12z}=\frac{2(z^2-4)}{6z(z+2)}=\frac{(z-2)(z+2)}{3(z+2)}=\frac{z-2}{3};$

в) $\frac{3a^2-6ab+3b^2}{6a-6b}=\frac{3(a^2-2ab+b^2)}{6(a-b)}=\frac{(a-b)^2}{2(a-b)}=\frac{a-b}{2};$

г) $\frac{an+3a}{an^2+6an+9a}=\frac{a(n+3)}{a(n^2+6a+9)}=\frac{(n+3)}{(n+3)^2}=\frac{1}{n+3};$

д) $\frac{p^3-p}{p^2-p}=\frac{p(p^2-p)}{(p^2-p)}=p;$

е) $\frac{a^2b+ab^2}{a^3b-ab^3}=\frac{ab(a+b)}{ab(a^2-b^2)}=\frac{(a+b)}{(a+b)(a-b)}=\frac{1}{a-b};$

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