a) $\frac{a}{a^2-9}-\frac{3}{9-a^2}=\frac{a}{a^2-9}+\frac{3}{a^2-9}=\frac{a+3}{a^2-9}=\frac{a+3}{(a+3)(a-3)}=\frac{1}{a-3}$.\newline b) $\frac{a^2}{a-1}+\frac{1}{1-a}=\frac{a^2}{a-1}-\frac{1}{a-1}=\frac{a^2-1}{a-1}=\frac{(a-1)(a+1)}{a-1}=a+1$.\newline c) $\frac{10 a-12}{2 a-6}+\frac{6 a}{6-2 a}=\frac{10 a-12}{2 a-6}-\frac{6 a}{2 a-6}=\frac{10 a-12-6 a}{2 a-6}=$ $=\frac{4 a-12}{2 a-6}=\frac{4(a-3)}{2(a-3)}=2$\newline d) $\frac{b-16}{2 b-12}-\frac{-3 b+8}{2 b-12}=\frac{b-16-(-3 b+8)}{2 b-12}=\frac{b-16+3 b-8}{2 b-12}=\frac{4 b-24}{2 b-12}=2$.

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Математика - 8

Задание 19 стр 106

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a) $\frac{a}{a^{2} - 9} - \frac{3}{9 - a^{2}} = \frac{a}{a^{2} - 9} + \frac{3}{a^{2} - 9} = \frac{a + 3}{a^{2} - 9} = \frac{a + 3}{(a + 3) (a - 3)} = \frac{1}{a - 3}$ .
b) $\frac{a^{2}}{a - 1} + \frac{1}{1 - a} = \frac{a^{2}}{a - 1} - \frac{1}{a - 1} = \frac{a^{2} - 1}{a - 1} = \frac{(a - 1) (a + 1)}{a - 1} = a + 1$ .
c) $\frac{10 a - 12}{2 a - 6} + \frac{6 a}{6 - 2 a} = \frac{10 a - 12}{2 a - 6} - \frac{6 a}{2 a - 6} = \frac{10 a - 12 - 6 a}{2 a - 6} =$ $= \frac{4 a - 12}{2 a - 6} = \frac{4 (a - 3)}{2 (a - 3)} = 2$
d) $\frac{b - 16}{2 b - 12} - \frac{- 3 b + 8}{2 b - 12} = \frac{b - 16 - (- 3 b + 8)}{2 b - 12} = \frac{b - 16 + 3 b - 8}{2 b - 12} = \frac{4 b - 24}{2 b - 12} = 2$ .

gimli

2023-01-12 14:22

Задание 19 стр 106

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