Question

Помогите пожалуйста
Найдите значение выражения:$\frac{3x-9x^{2} }{x^{2} +6x+9} :\frac{1-9x^{2} }{x^{2}-9 }$ при x=1

Answer 1

$\displaystyle\bf\\\frac{3x-9x^{2} }{x^{2} +6x+9} :\frac{1-9x^{2} }{x^{2} -9} =\frac{3x\cdot(1-3x)}{(x+3)^{2} } :\frac{(1-3x)\cdot(1+3x)}{(x-3)\cdot(x+3)} =\\\\\\=\frac{3x\cdot(1-3x)}{(x+3)^{2} } \cdot\frac{(x-3)\cdot(x+3)}{(1-3x)\cdot(1+3x)} =\frac{3x\cdot(x-3)}{(x+3)\cdot(1+3x)} \\\\\\x=1\\\\\\\frac{3x\cdot(x-3)}{(x+3)\cdot(1+3x)} =\frac{3\cdot 1\cdot(1-3)}{(1+3)\cdot(1+3\cdot 1)} =\\\\\\=\frac{3\cdot(-2)}{4\cdot 4} =\frac{-6}{16}=- 0,375$