ĐẶT: T= \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{\frac{ax^3}{x}+\frac{by^3}{y}+\frac{cz^3}{z}}=\sqrt[3]{ax^3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}=x\sqrt[3]{a}\)
\(\Rightarrow\sqrt[3]{a}=\frac{T}{x}\)
tuowng tự ta đc \(\sqrt[3]{b}=\frac{T}{y};\sqrt[3]{c}=\frac{T}{z}\)
\(\Rightarrow\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\frac{T}{x}+\frac{T}{y}+\frac{T}{z}=T\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=T\left(dpcm\right)\)

cám ơn  bạn nha!

Đặt:  \(ax^3=by^3=cz^3=k\)    và do    \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)  nên

\(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{\frac{ax^3}{x}+\frac{by^3}{y}+\frac{cz^3}{z}}\)

\(=\sqrt[3]{k\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}=\sqrt[3]{k}\)

\(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{\frac{k}{x^3}}+\sqrt[3]{\frac{k}{y^3}}+\sqrt[3]{\frac{k}{z^3}}\)

\(=\sqrt[3]{k}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\sqrt[3]{k}\)

suy ra:đpcm

Có: A= \(\sqrt[3]{ax^2+by^2+cz^2}\) = \(\sqrt[3]{\frac{ax^3}{x}+\frac{by^3}{y}+\frac{cz^3}{z}}\) = \(\sqrt[3]{ax^3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\) 

= \(\sqrt[3]{ax^3}\) = \(\sqrt[3]{a}x\) =>\(\sqrt[3]{a}\) =\(\frac{A}{x}\)

Tương tự : \(\sqrt[3]{b}=\frac{A}{y}\)   ,    \(\sqrt[3]{c}=\frac{A}{z}\)

=> \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\) = \(\frac{A}{x}+\frac{A}{y}+\frac{A}{z}\) = A \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) = A

hay \(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\) = \(\sqrt[3]{ax^2+by^2+cz^2}\)

Đây mà là ngữ văn lớp 1 á?

ngữ văn ko phải toán ko giải dc với đây là toán lớp 6 nha

đặt \(ax^3=by^3=cz^3=k^3\) thì \(a=\frac{k^3}{x^3};b=\frac{k^3}{y^3};c=\frac{k^3}{z^3}\)

\(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\frac{k}{x}+\frac{k}{y}+\frac{k}{z}=k\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=k\)

Mặt khác : \(ax^2+by^2+cz^2=\frac{ax^3}{x}+\frac{by^3}{y}+\frac{cz^3}{z}=\frac{k^3}{x}+\frac{k^3}{y}+\frac{k^3}{z}=k^3\)

\(\Rightarrow\sqrt[3]{ax^2+by^2+cz^2}=k\)

Do đó , ta có đpcm

\(\sin90\)

Đặt \(ax^3=by^3=cz^3=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{k}{x^3}\\b=\frac{k}{y^3}\\c=\frac{k}{z^3}\end{matrix}\right.\)

Thay vào VT ta được :

\(VT=\sqrt[3]{x^2\cdot\frac{k}{x^3}+y^2\cdot\frac{k}{y^3}+z\cdot\frac{k}{z^3}}=\sqrt[3]{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}\)

\(=\sqrt[3]{k\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}=\sqrt[3]{k}\) (1)

Thay vào VP ta được :

\(VP=\sqrt[3]{\frac{k}{x^3}}+\sqrt[3]{\frac{k}{y^3}}+\sqrt[3]{\frac{k}{z^3}}=\frac{\sqrt[3]{k}}{x}+\frac{\sqrt[3]{k}}{y}+\frac{\sqrt[3]{k}}{z}=\sqrt[3]{k}\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=\sqrt[3]{k}\) (2)

Từ (1) và (2) \(\Rightarrow VT=VP\)

Ta có đpcm.

Ta có: \(ax^3+by^3+cz^3=\frac{ax^3}{x}+\frac{by^3}{y}+\frac{cz^3}{z}\)

mà \(ax^3=by^3=cz^3\)

\(\Rightarrow ax^3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=ax^3\)

\(\Rightarrow\sqrt[3]{ax^3+by^3+cz^3}=x\sqrt[3]{a}\\ \Leftrightarrow\frac{\sqrt[3]{ax^3+by^3+cz^3}}{x}=\sqrt[3]{a}\\ \Leftrightarrow\sqrt[3]{ax^3+by^3+cz^3}.\frac{1}{x}=\sqrt[3]{a}\)

Tương tự, ta có:

\(\sqrt[3]{ax^3+by^3+cz^3}.\frac{1}{y}=\sqrt[3]{b}\)

\(\sqrt[3]{ax^3+by^3+cz^3}.\frac{1}{z}=\sqrt[3]{c}\)

Cộng vế theo vế các đẳng thức, ta có:

\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\sqrt[3]{ax^3+by^3+cz^3}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\\ =\sqrt[3]{ax^3+by^3+cz^3}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\left(đpcm\right)\)

Chúc bạn học tốt!

\(ax^3=by^3=cz^3\Rightarrow\frac{ax^2}{\frac{1}{x}}=\frac{by^2}{\frac{1}{y}}=\frac{cz^2}{\frac{1}{z}}=\frac{ax^2+by^2+cz^2}{1}\)

=> \(ax^2+by^2+cz^2=ax^3+by^3+cz^3\)

=> \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{ax^3}=\sqrt[3]{by^3}=\sqrt[3]{cz^3}=x\sqrt[3]{a}=y\sqrt[3]{b}=z\sqrt[3]{c}\) (1)

=> \(\frac{\sqrt[3]{a}}{\frac{1}{x}}=\frac{\sqrt[3]{b}}{\frac{1}{y}}=\frac{\sqrt[3]{c}}{\frac{1}{z}}=\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)   (2)

Tu (1) va (2) ta co dpcm

Chuc bn hoc tot !!!

Đặt \(ax^3=by^3=cz^3=k\).

Khi đó ta có:

\(VT=\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{\dfrac{k}{x}+\dfrac{k}{y}+\dfrac{k}{z}}=\sqrt[3]{k\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\sqrt[3]{k}\).

\(VP=\sqrt[3]{\dfrac{k}{x^3}}+\sqrt[3]{\dfrac{k}{y^3}}+\sqrt[3]{\dfrac{k}{z^3}}=\sqrt[3]{k}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\sqrt[3]{k}\).

Từ đó ta có đpcm.

Ta có: ax³ = \(\dfrac{ax^2}{\dfrac{1}{x}}\)

Tương tự ta có: ax³ = by³ = cz³ 

hay \(\dfrac{ax^2}{\dfrac{1}{x}}=\dfrac{by^2}{\dfrac{1}{y}}=\dfrac{cz^2}{\dfrac{1}{z}}=\dfrac{ax^2+by^2+cz^2}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\) = ax² + by² + cz² (T/c dãy tỉ số bằng nhau)

\(\Rightarrow\) \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{ax^3}=\sqrt[3]{by^3}=\sqrt[3]{cz^3}\)

= \(\dfrac{\sqrt[3]{a}}{\dfrac{1}{x}}=\dfrac{\sqrt[3]{b}}{\dfrac{1}{y}}=\dfrac{\sqrt[3]{c}}{\dfrac{1}{z}}=\dfrac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)  (đpcm)

Chúc bn học tốt!