a) $\frac{1}{x^2+x y}+\frac{1}{x y+y^0}=\frac{1}{x(x+y)}+\frac{1}{y(x+y)}=\frac{y+x}{x y(x+y)}=\frac{1}{x y}$.\newline b) $\frac{1}{x^2-x y}-\frac{1}{x y-y^2}=\frac{1}{x(x-y)}-\frac{1}{y(x-y)}=\frac{y-x}{x y(x-y)}=-\frac{1}{x y}$.\newline c) $\frac{a+6}{a^2-4}-\frac{2}{a^2+2 a}=\frac{a+6}{(a-2)(a+2)}-\frac{2}{a(a+2)}=\frac{a(a+6)-2(a-2)}{2 a(a+2)(a-2)}=\frac{a^2+6 a-2 a+4}{2 a(a+2)(a-2)}=\frac{a^2+4 a+4}{2 a(a+2)(a-2)}=\frac{(a+2)^2}{2 a(a+2)(a-2)}=\frac{a+2}{2 a(a-2)}=\frac{a+2}{2 a^2-4 a} \text {. } $

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Математика - 8

Задание 16 стр 106

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a) $\frac{1}{x^{2} + xy} + \frac{1}{xy + y^{0}} = \frac{1}{x (x + y)} + \frac{1}{y (x + y)} = \frac{y + x}{xy (x + y)} = \frac{1}{xy}$ .
b) $\frac{1}{x^{2} - xy} - \frac{1}{xy - y^{2}} = \frac{1}{x (x - y)} - \frac{1}{y (x - y)} = \frac{y - x}{xy (x - y)} = - \frac{1}{xy}$ .
c) $\frac{a + 6}{a^{2} - 4} - \frac{2}{a^{2} + 2 a} = \frac{a + 6}{(a - 2) (a + 2)} - \frac{2}{a (a + 2)} = \frac{a (a + 6) - 2 (a - 2)}{2 a (a + 2) (a - 2)} = \frac{a^{2} + 6 a - 2 a + 4}{2 a (a + 2) (a - 2)} = \frac{a^{2} + 4 a + 4}{2 a (a + 2) (a - 2)} = \frac{{(a + 2)}^{2}}{2 a (a + 2) (a - 2)} = \frac{a + 2}{2 a (a - 2)} = \frac{a + 2}{2 a^{2} - 4 a} .$

gimli

2023-01-12 13:57

Задание 16 стр 106

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