Question

Решите,,не хватает времени сделать эти 3 номерa! №1 разложить на множители: 1)ас⁴-с⁴-ас²+с²= 2)х³у-ху-х³+х= №2 сократить дробь: 1)3х²-7х+2/2-6х= 2)5х²-12х+4/6-15х= №3 выр-ние: 1) (2m/2m+n - 4m²/4m²+4mn+n²) : ( 2m/4m²-n² + 1/n-2m) = 2)( x²/x+y - x³/x²+y²+2xy ): (x/x+y + x²/y²-x²)=

Answer 1

$ac^4-c^4-ac^2+c^2=\\ (ac^4-c^4)-(ac^2-c^2)=\\ c^4(a-1)-c^2(a-1)=\\ (c^4-c^2)(a-1)=\\ c^2(c^2-1)(a-1)=\\ c^2(c-1)(c+1)(a-1)$

$x^3y-xy-x^3+x=\\ (x^3y-xy)-(x^3-x)=\\ y(x^3-x)-(x^3-x)=\\ (y-1)(x^3-x)=\\ (y-1)x(x^2-1)=\\ (y-1)x(x-1)(x+1)$

$\frac{3x^2-7x+2}{2-6x}=\\ \frac{3x^2-6x-x+2}{2(1-3x)}=\\ \frac{-3x(2-x)+1(2-x)}{2(1-3x)}=\\ \frac{-3x(2-x)+1(2-x)}{2(1-3x)}=\\ \frac{(1-3x)(2-x)}{2(1-3x)}=\\ \frac{2-x}{2}$

$\frac{5x^2-12x+4}{6-15x}=\\ \frac{5x^2-2x-10x+4}{3(2-5x)}=\\ \frac{-x(2-5x)+2(2-5x)}{3(2-5x)}=\\ \frac{(2-x)(2-5x)}{3(2-5x)}=\\ \frac{2-x}{3}$

$(\frac{2m}{2m+n}-\frac{4m^2}{4m^2+4mn+n^2}:(\frac{2m}{4m^2-n^2}+\frac{1}{n-2m})=\\ (\frac{2m}{2m+n}-\frac{4m^2}{(2m+n)^2}:(\frac{2m}{(2m-n)(2m+n)}-\frac{1}{2m-n})=\\ (\frac{2m(2m+n)}{(2m+n)^2}-\frac{4m^2}{(2m+n)^2}:(\frac{2m}{(2m-n)(2m+n)}-\frac{1*(2m+n)}{(2m-n)(2m+n)})=\\ (\frac{4m^2+2mn}{(2m+n)^2}-\frac{4m^2}{(2m+n)^2}:(\frac{2m}{(2m-n)(2m+n)}-\frac{2m+n}{(2m-n)(2m+n)})$

$\frac{4m^2+2mn-4m^2}{(2m+n)^2}:\frac{2m-2m-n}{(2m-n)(2m+n)}=\\ \frac{2mn}{(2m+n)^2}:\frac{-n}{(2m-n)(2m+n)}=\\ \frac{2mn}{(2m+n)^2}*\frac{(2m-n)(2m+n)}{-n}=\\ \frac{2mn(2m-n)(2m+n)}{(2m+n)^2(-n)}=\\ \frac{2m(n-2m)}{(2m+n)}=\\$

$(\frac{x^2}{x+y}-\frac{x^3}{x^2+y^2+2xy}):(\frac{x}{x+y}+\frac{x^2}{y^2-x^2})=\\$

$(\frac{x^2}{x+y}-\frac{x^3}{(x+y)^2}):(\frac{x}{x+y}-\frac{x^2}{(x-y)(x+y)})=\\ (\frac{x^2(x+y)}{(x+y)^2}-\frac{x^3}{(x+y)^2}):(\frac{x(x-y)}{(x+y)(x-y)}-\frac{x^2}{(x-y)(x+y)})=\\ \frac{x^2(x+y)-x^3}{(x+y)^2}:\frac{x(x-y)-x^2}{(x-y)(x+y)}=\\ \frac{x^3+xy-x^3}{(x+y)^2}:\frac{x^2-xy-x^2}{(x-y)(x+y)}=\\ \frac{xy}{(x+y)^2}:\frac{-xy}{(x-y)(x+y)}=\\ \frac{xy}{(x+y)^2}*\frac{(x-y)(x+y)}{-xy}=\\ \frac{xy(x-y)(x+y)}{-xy(x+y)^2}=\\ \frac{y-x}{y+x}$

Answer 2

№1

 c^2[(ac^2-a)-(c^2-1)]=c^2[a(c^2-1)-(c^2-1)]=c^2(c^2-1)(a-1)=c^2(c-1)(c+1)(a-1)

x(x^2y-y-x^2+1) =x[y(x^2-1)-(x^2-1)]=x(x-1)(x+1)(y-1)

№2

Разложим числитель на множители: (x-2)(3*x-1)

(x-2)(3*x-1)/2(1-3*x)=-(x-2)(1-3*x)/2(1-3*x)=(2-x)/2

(x-2)(5x-2)/(3(2-5x)=-(x-2)(2-5x)/(3(2-5x)=(2-x)/3

№3

Упростим сначала числитель:

(8m^3+8m^2n+2mn^2-8m^3-4m^2n)/(8m^3+2m^2n+2mn^2+4m^2n+mn^2+n^3)=

=(2mn(2m+n))/(2m+n)^3=2mn/(2m+n)^2

Теперь знаменатель:

(2mn^2-4m^2+4m^2-n^2)/(4m^2n-8m^3-n^3+2mn^2)=-n(2m-n)/(2m-n)^3=

=-n/(2m-n)^2

Соединяем:

-(2nm(2m-n)^2)/((2m+n)^2)*n)=-(2m(2m-n)^2/(2m+n)^2

Числитель: 

x^2/(x+y)-x^3/(x+y)^2=(x^3+x^2y-x^3)/(x+y)^2=x^2y/(x+y)^2

Знаменатель:

x/(x+y)+x^2/(y+x)(y-x)=(xy-x^2+x^2)/(y+x)(y-x)=xy/((y+x)(y-x))

Собираем:x^2y*(y+x)(y-x)/((x+y)^2*xy)=x(y-x)/(x+y)