Question

Срочно!!!!!!!!!!!!!!!!!!!!!!!​

Answer 1

-фотографии-

К сожалению,8 не сделал,так как в таком плохо разбираюсь

Answer 2

7. Спростити вираз:

$(\frac{1-y^2}{1-y}-2y)\div\frac{y^2-2y+1}{y+1}+\frac{2y}{y-1}=\\\\=(\frac{(1-y)(1+y)}{1-y}-2y)\div\frac{y^2-2y+1}{y+1}+\frac{2y}{y-1}=\\\\=(1+y-2y)\div\frac{y^2-2y+1}{y+1}+\frac{2y}{y-1}=\\\\=(1-y)\div\frac{y^2-2y+1}{y+1}+\frac{2y}{y-1}=\\\\=(1-y)*\frac{y+1}{y^2-2y+1}+\frac{2y}{y-1}=\\\\=\frac{(1-y)(y+1)}{(y-1)(y-1)}+\frac{2y}{y-1}=\\\\=\frac{-(y-1)(y+1)}{(y-1)(y-1)}+\frac{2y}{y-1}=\\\\=\frac{-(y+1)}{y-1}+\frac{2y}{y-1}=\\\\=\frac{2y-(y+1)}{y-1}=\\\\=\frac{y-1}{y-1}=\\\\=1.$

8. Обчисліть значення виразу:

$\frac{x}{y}=\frac{1}{2}\\\\y=2x\\\\\frac{4x^2-9y^2+2xy}{3y^2+16x^2}=\frac{(2x)^2-9y^2+(2x)*y}{3y^2+4*(2x)^2}=\\\\=\frac{y^2-9y^2+y*y}{3y^2+4y^2}=\frac{-7y^2}{7y^2}=-1.$

9. Доведіть тотожність:

$\frac{8x^2+14x+3}{8x^3-1}-\frac{1-4x}{4x^2+2x+1}-\frac{3}{2x-1}=\frac{1}{2x-1}\\\\\frac{8x^2+14x+3}{(2x)^3-1^3}-\frac{1-4x}{4x^2+2x+1}-\frac{3}{2x-1}=\frac{1}{2x-1}\\\\\frac{8x^2+14x+3}{(2x-1)((2x)^2+(2x)*1+1^2)}-\frac{1-4x}{4x^2+2x+1}-\frac{3}{2x-1}=\frac{1}{2x-1}\\\\\frac{8x^2+14x+3}{(2x-1)(4x^2+2x+1)}-\frac{1-4x}{4x^2+2x+1}-\frac{3}{2x-1}=\frac{1}{2x-1}\\\\\frac{8x^2+14x+3}{(2x-1)(4x^2+2x+1)}-\frac{(1-4x)(2x-1)}{(4x^2+2x+1)(2x-1)}-\frac{3(4x^2+2x+1)}{(2x-1)(4x^2+2x+1)}=\frac{1}{2x-1}$

$\frac{8x^2+14x+3-(1-4x)(2x-1)-3(4x^2+2x+1)}{(2x-1)(4x^2+2x+1)}=\frac{1}{2x-1}\\\\\frac{8x^2+14x+3-(-8x^2+6x-1)-(12x^2+6x+3)}{(2x-1)(4x^2+2x+1)}=\frac{1}{2x-1}\\\\\frac{8x^2+14x+3+8x^2-6x+1-12x^2-6x-3}{(2x-1)(4x^2+2x+1)}=\frac{1}{2x-1}\\\\\frac{8x^2+14x+3+8x^2-6x+1-12x^2-6x-3}{(2x-1)(4x^2+2x+1)}=\frac{1}{2x-1}\\\\\frac{4x^2+2x+1}{(2x-1)(4x^2+2x+1)}=\frac{1}{2x-1}\\\\\frac{1}{2x-1}=\frac{1}{2x-1}.$