Question

Найди значение выражения в^2-16в+12)/в^3+8+3в+2/в^2-2в+4. При в=-2,1

Answer 1

$\displaystyle\bf\\\frac{b^{2} -16b+12}{b^{3}+8 } +\frac{3b+2}{b^{2} -2b+4} =\frac{b^{2} -16b+12}{(b+2)(b^{2} -2b+4)} +\frac{3b+2}{b^{2} -2b+4} =\\\\\\=\frac{b^{2} -16b+12+(3b+2)(b+2)}{(b+2)(b^{2} -2b+4)} =\frac{b^{2}-16b+12+3b^{2} +6b+2b+4 }{(b+2)(b^{2} -2b+4)} =\\\\\\=\frac{4b^{2}-8b+16 }{(b+2)(b^{2} -2b+4)} =\frac{4(b^{2}-2b+4) }{(b+2)(b^{2}-2b+4) }=\frac{4}{b+2} \\\\\\b=-2,1\\\\\\\frac{4}{b+2} =\frac{4}{-2,1+2} =-\frac{4}{0,1}=-40$

Answer 2

Ответ:

19023 правильно

Объяснение:

правильно