Chứng minh đẳng thức $\sqrt[3]{ax^2+by^2+cz^2}$ = $\sqrt[3]{a}$ + $\sqrt[3]{b}$ + $\sqrt[3]{c}$ Biết rẳng $ax^{3}$=$by^{3}$=$cz^{3}$ và $\frac{1}{

Đặt `ax^3=by^3=cz^3=k`

$⇒\begin{cases}x^3=\dfrac{k}{a}⇒\dfrac{1}{x^3}={\dfrac{a}{k}}⇒\dfrac{1}{x}=\sqrt[3]{\dfrac{a}{k}}\\y^3=\dfrac{k}{b}⇒\dfrac{1}{y^3}={\dfrac{b}{k}}⇒\dfrac{1}{y}=\sqrt[3]{\dfrac{b}{k}}\\z^3=\dfrac{k}{c}⇒\dfrac{1}{z^3}={\dfrac{c}{k}}⇒\dfrac{1}{z}=\sqrt[3]{\dfrac{c}{k}}\end{cases}$   

Ta cộng tổng:

`1=1/x+1/y+1/z=`$\sqrt[3]{\dfrac{a}{k}}+\sqrt[3]{\dfrac{b}{k}}+\sqrt[3]{\dfrac{c}{k}}={\dfrac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\sqrt[3]{k}}}⇒\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{k}$ 

Ta cần chứng minh: $\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{k}$

`ax^3=by^3=cz^3=k`

$⇒\begin{cases}a=\dfrac{k}{x^3}\\b=\dfrac{k}{y^3}\\c=\dfrac{k}{z^3}\end{cases}$   

`⇒` $\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{\dfrac{k}{x^3}.x^2+\dfrac{k}{y^3}.y^2+\dfrac{k}{z^3}.z^2}=\sqrt[3]{\dfrac{k}{x}+\dfrac{k}{y}+\dfrac{k}{z}}=\sqrt[3]{k.(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}})=\sqrt[3]{k.1}=\sqrt[3]{k}$

Vậy ta có điều phải chứng minh: $\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}.$