Question

№1)дробь: х+4/х2+5х+4= сократите х2-3х-18/х-6 х2-4/х2-5х+6 х2+10х+21/х2-9 № 2) дробь: 10х2-6х+5хв-3в/5х2-8х+3 сократите 2х2-10х-хв+5х/3х2-14х-5 сократите зарание большое

Answer 1

Задание №1

$\frac{x+4}{x^{2}+5x+4}=\frac{(x+4)}{(x+1)(x+4)}=\frac{1}{x+1}$

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$\frac{x^{2}-3x-18}{x-6}=\frac{(x-6)(x+3)}{(x-6)}=x+3$

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

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$\frac{x^{2}+10x+21}{x^{2}-9}=\frac{(x+3)(x+7)}{(x+3)(x-3)}=\frac{x+7}{x-3}$

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Задание № 2

$\frac{10x^{2}-6x+5xb-3b}{5x^{2}-8x+3}=\frac{(10x^{2}-6x)+(5xb-3b)}{(5x-3)(x-1)}=\frac{2x(5x-3)+b(5x-3)}{(5x-3)(x-1)}=\frac{(5x-3)(2x+b)}{(5x-3)(x-1)}=\frac{2x+b}{x-1}$

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$\frac{2x^{2}-10x-xb+5b}{3x^{2}-14x-5}=\frac{(2x^{2}-10x)+(-xb+5b)}{(x-5)(3x+1)}=\frac{(2x^{2}-10x)-(xb-5b)}{(x-5)(3x+1)}=\frac{2x(x-5)-b(x-5)}{(x-5)(3x+1)}=\frac{(x-5)(2x-b)}{(x-5)(3x+1)}=\frac{2x-b}{3x+1}$