Предмет: Математика
Автор: pikosha02
Вопрос задан 1 год назад

Срочнооооо решите тождество !!!!!!!!!!!!

Приложения:

Ответы

Ответ дал: Аноним

$1)\\A.\;\frac{x^{\frac3p}-x^{\frac3q}}{\left(x^{\frac1p}+x^{\frac1q}\right)^2-2x^{\frac1q}\left(x^{\frac1q}+x^{\frac1p}\right)}=\frac{x^{\frac3p}-x^{\frac3q}}{x^{\frac2p}+2x^{\frac1p}x^{\frac1q}+x^{\frac2q}-2x^{\frac2q}-2x^{\frac1p}x^{\frac1q}\right)}=\frac{x^{\frac3p}-x^{\frac3q}}{x^\frac2p-x^{\frac2q}}=$

$=\frac{\left(x^{\frac1p}-x^{\frac1q}\right)\left(x^\frac2p+x^\frac1px^\frac1q+x^\frac2q\right)}{\left(x^\frac1p-x^\frac1q\right)\left(x^\frac1p+x^\frac1q\right)}=\frac{x^\frac2p+x^\frac1px^\frac1q+x^\frac2q}{x^\frac1p+x^\frac1q}$

$B.\;\frac{x^\frac1p}{x^{\frac{q-p}{pq}}+1}=\frac{x^\frac1p}{x^{\frac q{pq}-\frac p{pq}}+1}=\frac{x^\frac1p}{\frac{x^\frac1p}{x^\frac1q}+1}=\frac{x^\frac1px^\frac1q}{x^\frac1p+x^\frac1q}$

$A+B=\frac{x^\frac2p+x^\frac1px^\frac1q+x^\frac2q}{x^\frac1p+x^\frac1q}+\frac{x^\frac1px^\frac1q}{x^\frac1p+x^\frac1q}=\frac{x^\frac2p+2x^\frac1px^\frac1q+x^\frac2q}{x^\frac1p+x^\frac1q}=\frac{\left(x^\frac1p+x^\frac1q\right)^2}{x^\frac1p+x^\frac1q}=\\\\=x^\frac1p+x^\frac1q=\sqrt[p]x+\sqrt[q]x$

$2)\\\left(\frac{9-4a^{-2}}{3a^{-\frac12}+2a^{-\frac32}}-\frac{1+a^{-1}-6a^{-2}}{a^{-\frac12}+3a^{-\frac32}}\right)^4-16a^2=\\\\=\left(\frac{\left(3-2a^{-1}\right)\left(3+2a^{-1}\right)}{a^{-\frac12}\left(3+2a^{-1}\right)}-\frac{1-2a^{-1}+3a^{-1}-6a^{-2}}{a^{-\frac12}\left(1+3a^{-1}\right)}\right)^4-16a^2=\\\\=\left(\frac{3-2a^{-1}}{a^{-\frac12}}-\frac{1-2a^{-1}+3a^{-1}\left(1-2a^{-1}\right)}{a^{-\frac12}\left(1+3a^{-1}\right)}\right)^4-16a^2=$

$=\left(\frac{3-2a^{-1}}{a^{-\frac12}}-\frac{\left(1-2a^{-1}\right)\left(1+3a^{-1}\right)}{a^{-\frac12}\left(1+3a^{-1}\right)}\right)^4-16a^2=\left(\frac{3-2a^{-1}}{a^{-\frac12}}-\frac{1-2a^{-1}}{a^{-\frac12}}\right)^4-16a^2=\\\\=\left(\frac{3-2a^{-1}-1+2a^{-1}}{a^{-\frac12}}\right)^4-16a^2=\left(\frac2{a^{-\frac12}}\right)^4-16a^2=\left(2a^{\frac12}\right)^4-16a^2=\\\\=16a^2-16a^2=0$