Question

помогите упростить пожалуйста (подробно)

Answer 1

Упростить

$left (frac{1}{a+sqrt{2}}-frac{a^2+2}{a^3+2sqrt{2}} right )^{-1} cdot left(frac{a}{2}-frac{1}{sqrt{2}}+frac{1}{a} right )^{-1} cdot frac{sqrt{2}}{a+sqrt{2}}$

Решение

$a^3+2sqrt{2}=a^3+(sqrt{2})^3=(a+sqrt{2})(a^2-sqrt{2}a+2)$

$frac{1}{a+sqrt{2}}-frac{a^2+2}{a^3+2sqrt{2}}=frac{1}{a+sqrt{2}}-frac{a^2+2}{(a+sqrt{2})(a^2-sqrt{2}a+2)}=frac{a^2-sqrt{2}a+2-a^2-2}{(a+sqrt{2})(a^2-sqrt{2}a+2)}=frac{-sqrt{2}a}{(a+sqrt{2})(a^2-sqrt{2}a+2)}$

$left (frac{1}{a+sqrt{2}}-frac{a^2+2}{a^3+2sqrt{2}} right )^{-1}=left (frac{-sqrt{2}a}{(a+sqrt{2})(a^2-sqrt{2}a+2)} right )^{-1}=-frac{(a+sqrt{2})(a^2-sqrt{2}a+2)}{sqrt{2}a}$

$left (frac{a}{2}-frac{1}{sqrt{2}}+frac{1}{a} right )^{-1}=left (frac{a^2-sqrt{2}a+2}{2a} right )^{-1}=frac{2a}{a^2-sqrt{2}a+2}$

$left (frac{1}{a+sqrt{2}}-frac{a^2+2}{a^3+2sqrt{2}} right )^{-1} cdot left(frac{a}{2}-frac{1}{sqrt{2}}+frac{1}{a} right )^{-1}=-frac{(a+sqrt{2})(a^2-sqrt{2}a+2)}{sqrt{2}a} cdot frac{2a}{a^2-sqrt{2}a+2}=-sqrt{2}(a+sqrt{2})$

$left (frac{1}{a+sqrt{2}}-frac{a^2+2}{a^3+2sqrt{2}} right )^{-1} cdot left(frac{a}{2}-frac{1}{sqrt{2}}+frac{1}{a} right )^{-1}cdotfrac{sqrt{2}}{a+sqrt{2}}=-sqrt{2}(a+sqrt{2})cdotfrac{sqrt{2}}{a+sqrt{2}}=-2$

Ответ: -2

Answer 2

Спасибо))