Question

ФОТКА С ЗАДАНИЕ СРОЧНО ЗА ОТВЕТ 50 БАЛОВ

Answer 1

Ответ:

Объяснение:

1)$\frac{3d-c}{cd}-\frac{4c+3d}{cd}= \frac{3d-c-(4c+3d)}{cd} = \frac{3d-c-4c-3d}{cd} =\frac{-5c}{cd} = -\frac{5}{d}$

2)$\frac{a-8}{a^{2}-25 } +\frac{13}{a^{2}-25} = \frac{a-8+13}{a^{2}-25}= \frac{a-5}{(a-5)(a+5)} = a+5$

3) $\frac{3b+7}{3b} -\frac{b^{2}-5 }{b^{2} } = \frac{b(3b+7)}{3b^{2}} -\frac{3(b^{2}-5)}{3b^{2}} = \frac{b(3b+7)-3(b^{2} -5)}{3b^{2}} = \frac{3b^{2}+7b-3b^{2} +15 }{3b^{2} } = \frac{7b+15}{3b^{2} }$

4) $\frac{1}{3x+y} -\frac{1}{3x-y} = \frac{(3x-y)-(3x+y)}{(3x-y)(3x+y)} =\frac{-2y}{(3x-y)(3x+y)}$

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

$\frac{x-6y^{2} }{2y} +3y=\frac{x-6y^{2} }{2y}+\frac{6y^{2} }{2y} = \frac{x-6y^{2} +6y^{2} }{2y}= \frac{x}{2y} = \frac{-8}{2*0,1} = \frac{-8}{0,2} = -40$