Question

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Во вложениях  номер 241(4,5,6)    ещё сверху 3 и  4  и 243 полностью

Answer 1

$241.\ 4.;frac4{x+5};;frac3{7x}\ O.3.:;;7x(x+5)\ frac{28x}{7x(x+5)};;frac{3(x+5)}{7x(x+5)}\ 5.;frac a{2a+2};;frac3{a+1}\ frac a{2(a+1)};;frac3{a+1}\ O.3.:;;2(a+1)\ frac a{2a+2};;frac6{2(a+1)}\ 6.;frac{2x}{3x-3y};;frac{3y}{4x+4y}\ frac{2x}{3(x-y)};;frac{3y}{4(x+y)}\ O.3.:;;12(x+y)(x-y)=12(x^2-y^2)=12x^2-12y^2\ frac{8x(x+y)}{12x^2-12y^2};;frac{9y(x-y)}{12x^2-12y^2}$

$242.\<var>3.;;frac{x-1}{x^2-8x+16};;frac{x+2}{16-x^2}\ frac{x-1}{(x-4)^2};;frac{x+2}{(4-x)(4+x)}\ frac{x-1}{(4-x)^2};;frac{x+2}{(4-x)(4+x)}\ O.3.:;(4-x)^2(4+x)\ frac{(4+x)(x-1)}{(4-x)^2(4+x)};;frac{(4-x)(x+2)}{(4-x)^2(4+x)}\ 4.;;frac{2-x}{25+10x+x^2};;frac{1+x}{50-2x^2}\ frac{2-x}{(5+x)^2};;frac{1+x}{2(25-x^2)}\ frac{2-x}{(5+x)^2};;frac{1+x}{2(5-x)(5+x)}\ O.3.:;;2(5-x)(5+x)^2\ frac{2(5-x)(2-x)}{2(5-x)(5+x)^2};;frac{(5+x)(1+x)}{2(5-x)(5+x)^2}</var>$

$243.\ 1.;frac{a^2-ax}{a^2x-ax^2}=frac{a^2-x}{x(a^2-ax)}=frac1x\ 2.;frac{mn^4-cn^4}{cn^3-mn^3}=frac{-n(cn^3-mn^3)}{cn^3-mn^3}=-n\ 3.;frac{4p^2-16p^3}{12p^2-3p}=frac{-4p(4p^2-p)}{3(4p^2-p)}=-frac{4p}3\ 4.;frac{a^2-2a+1}{a^2-1}=frac{(a-1)^2}{(a-2)(a+2)}=frac{(a-2)}{(a+2)}\ 5.;frac{b^3+1}{b^2-b+1}=frac{(b+1)(b^2-b+1)}{b^2-b+1}=b+1\ 6.;frac{1-n^2}{1-n^3}=frac{(1-n)(1+n)}{(1-n) (1+n+n^2)}=frac{(1+n)}{1+n+n^2}\ 7.;frac{x^2-4x+16}{ax-4a}=frac{(x-4)^2}{a(x-4)}=frac{x-4}a$

$8.;frac{3a^2-ax}{9a^2-6ax+x^2}=frac{a(3a-x)}{(3a-x)^2}=frac a{3a-x}\ 9.;frac{a^3-a^2-a+1}{1-2a^2+a^4}=frac{-a^2(1-a)+(1-a)}{(1-a^2)^2}=frac{(1-a)(1-a^2)}{(1-a^2)^2}=frac{1-a}{(1-a)(1+a)}=frac1{1+a}$