Question

Срочно нужно выполнить действие

Answer 1

Объяснение:

$\frac{1}{x + 2} - \frac{x}{x {}^{2} - 2x + 4 } - \frac{12}{x {}^{3} + 8 } =\frac{x^{2}-2x+4}{\left(x+2\right)\left(x^{2}-2x+4\right)}-\frac{x\left(x+2\right)}{\left(x+2\right)\left(x^{2}-2x+4\right)}-\frac{12}{x^{3}+8} =\frac{x^{2}-2x+4-x\left(x+2\right)}{\left(x+2\right)\left(x^{2}-2x+4\right)}-\frac{12}{x^{3}+8} =\frac{x^{2}-2x+4-x^{2}-2x}{\left(x+2\right)\left(x^{2}-2x+4\right)}-\frac{12}{x^{3}+8} =\frac{-4x+4}{\left(x+2\right)\left(x^{2}-2x+4\right)}-\frac{12}{x^{3}+8} =\frac{-4x+4}{\left(x+2\right)\left(x^{2}-2x+4\right)}-\frac{12}{\left(x+2\right)\left(x^{2}-2x+4\right)} =\frac{-4x+4-12}{\left(x+2\right)\left(x^{2}-2x+4\right)} =\frac{-4x-8}{\left(x+2\right)\left(x^{2}-2x+4\right)} =\frac{4\left(-x-2\right)}{\left(x+2\right)\left(x^{2}-2x+4\right)} =\frac{-4\left(x+2\right)}{\left(x+2\right)\left(x^{2}-2x+4\right)} =\frac{-4}{x^{2}-2x+4} =-\frac{4}{x^{2}-2x+4}$