решите пожалуйста, нужно завтра!!. Created by Марьяна52. algebra-ru

решите пожалуйста, нужно завтра!!...

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1) x²y/x^8y^7 = 1/x^6y^6
2) - (3x^4y^4)/6x^7 = - y^4/2x³
3) (- 15x²y^7)/(- 9x^9y^8c) = 5/3x^7yc
4) (4x + 16)/x² - 16 = 4(x + 4)/(x - 4)(x + 4) = 4/x - 4
5) (x² - 14x + 49)/(x² - 49) = (x - 7)(x - 7)/ (x - 7)(x + 7) = (x - 7)/(x + 7)
6) (x^8 - x^6)/(x + 1) = x^6(x² - 1)/(x + 1) = x^6(x - 1)(x + 1)/(x + 1) =
= x^6(x - 1)
7) (2 - 2b²)/ (4b² - 8b + 4) = 2(1 - b²)/4(b - 1)(b - 1) = - (1 + b)/ 2(b - 1) =
(1 + b)/(1 - b)
8) (a² + 2ab + b²)/(a² -b²) = (a + b)(a + b)/(a + b)(a - b) = (a + b)/(a - b)
9) (12y³ - 8y²)/(2 - 3y) = 4y²(3y- 2)/(2 - 3y) = - 4y²

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sedinalana 1
$1/x^6y^5$
2
$-y^4/(2x^3)$
3
$-5/(3x^7yc)$
4
$4(x+4)/[(x-4)(x+4)]=4/(x-4)$
5
$(x-7)^2/[(x-7)(x+7)]=(x-7)/(x+7)$
6
$x^6(x^2-1)/(x+1)=x^6(x-1)(x+1)/(x+1)=x^6(x-1)$
7
$2(1-b^2)/[4(1-2b+b^2)]=(1-b)(1+b)/[2(1-b)^2]=(1+b)/[2(1-b$ $)]$
8
$(a+b)^2/[(a-b)(a+b)]=(a+b)/(a-b)$
9
$4y^2(3y-2)/[-(3y-2)]=-4y^2$

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