Question

Выполните вычитание: х^2/х^2-49 - х/х + 7

7b - 21b^2/3b + 4

выражение:

а - 18/2а-12 - а -6/2а + 12 + 50/ а^2 - 36

6с^3 + 3с/ с^3 - 1 - 3с^2/ с^2 + с + 1

Answer 1

$\dfrac{x^2}{x^2-49}-\dfrac{x}{x+7}=\dfrac{x^2}{(x-7)(x+7)}-\dfrac{x}{x+7}=\dfrac{x^2-x(x-7)}{(x+7)(x-7)}=\\ \\ \\ =\dfrac{x^2-x^2+7x}{(x+7)(x-7)}=\dfrac{7x}{x^2-49}$

$7b-\dfrac{21b^2}{3b+4}=\dfrac{7b(3b+4)-21b^2}{3b+4}=\dfrac{21b^2+28b-21b^2}{3b+4}=\dfrac{28b}{3b+4}$

Упростить выражение

$\dfrac{a-18}{2a-12}-\dfrac{a-6}{2a+12}+\dfrac{50}{a^2-36}=\dfrac{a-18}{2(a-6)}-\dfrac{(a-6)}{2(a+6)}+\dfrac{50}{(a-6)(a+6)}=\\ \\ \\ =\dfrac{(a-18)(a+6)-(a-6)^2+50\cdot 2}{2(a-6)(a+6)}=\dfrac{a^2-12a-108-a^2+12a-36+100}{2(a-6)(a+6)}=\\ \\ =-\dfrac{44}{2(a-6)(a+6)}=-\dfrac{22}{a^2-36}=\dfrac{22}{36-a^2}$

$\dfrac{6c^3+3c}{c^3-1}-\dfrac{3c^2}{c^2+c+1}=\dfrac{6c^3+3c}{(c-1)(c^2+c+1)}-\dfrac{3c^2}{c^2+c+1}=\\ \\ \\ =\dfrac{6c^3+3c-3c^2(c-1)}{(c-1)(c^2+c+1)}=\dfrac{6c^3+3c-3c^3+3c^2}{(c-1)(c^2+c+1)}=\\ \\\\ =\dfrac{3c^3+3c^2+3c}{(c-1)(c^2+c+1)}=\dfrac{3c(c^2+c+1)}{(c-1)(c^2+c+1)}=\dfrac{3c}{c-1}$