\(\Leftrightarrow30x^2+20y^2+15z^2=12x^2+12y^2+12z^2.\)

\(\Leftrightarrow18x^2+8y^2+3z^2=0\)(1)

\(x^2\ge0\Rightarrow18x^2\ge0\)

\(y^2\ge0\Rightarrow8y^2\ge0\)

\(z^2\ge0\Rightarrow3z^2\ge0\)

=> (1) = 0  khi \(18x^2=8y^2=3z^2=0\Rightarrow x=y=z=0\)

\(pt\Leftrightarrow\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}-\frac{x^2+y^2+z^2}{5}=0\)

\(\Leftrightarrow\left(\frac{x^2}{2}-\frac{x^2}{5}\right)+\left(\frac{y^2}{3}-\frac{y^2}{5}\right)+\left(\frac{z^2}{4}-\frac{z^2}{5}\right)=0\)

\(\Leftrightarrow\frac{3}{10}x^2+\frac{2}{15}y^2+\frac{1}{20}z^2=0\)

Ta thấy \(VT\ge0\forall x;y;z\) nên để dấu "=" xảy ra \(\Leftrightarrow x=y=z=0\)

(6x²+4y²+3z²)/12 = (x²+y²+z²)/5

30x²+20y²+15z²=12x²+12y²+12z²

18x²+8y²+3z²=0

=> x=y=z=0

vì x²;y²;z² > hoặc = 0

\(\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}=\frac{x^2+y^2+z^2}{5}\)

\(\Leftrightarrow\frac{x^2}{2}-\frac{x^2}{5}+\frac{y^2}{3}-\frac{y^2}{5}+\frac{z^2}{4}-\frac{z^2}{5}=0\)

\(\Leftrightarrow x^2\left(\frac{1}{2}-\frac{1}{5}\right)+y^2\left(\frac{1}{3}-\frac{1}{5}\right)+z^2\left(\frac{1}{4}-\frac{1}{5}\right)=0\)

\(\Leftrightarrow x=y=z=0\)

=>\(\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}=\frac{x^2}{5}+\frac{y^2}{5}+\frac{z^2}{5}\)

=>\(\left(\frac{x^2}{2}-\frac{x^2}{5}\right)+\left(\frac{y^2}{3}-\frac{y^2}{5}\right)+\left(\frac{z^2}{4}-\frac{z^2}{5}\right)=0\)

mà x²,y²,z² \(\ge\)0

=>\(\frac{x^2}{2},\frac{y^2}{3},\frac{z^2}{4},\frac{x^2}{5},\frac{y^2}{5},\frac{z^2}{5}\ge0\)

\(\Rightarrow\left(\frac{x^2}{2}-\frac{x^2}{5}\right)\ge0,\left(\frac{y^2}{3}-\frac{y^2}{5}\right)\ge0,\left(\frac{z^2}{4}-\frac{z^2}{5}\right)\ge0\)

Dấu bằng xảy ra khi:

\(\frac{x^2}{2}=\frac{x^2}{5},\frac{y^2}{3}=\frac{y^2}{5},\frac{z^2}{4}=\frac{z^2}{5}\)

\(\Rightarrow\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)

Đặt \(\frac{x^2}{2}=\frac{y^2}{3}=\frac{z^2}{4}=\frac{x^2+y^2+z^2}{5}=k\)(k >= 0)

=> \(x^2=2k;y^2=3k;z^2=4k;x^2+y^2+z^2=5k\)

=> \(x^2+y^2+z^2=2k+3k+4k=9k=5k\)

=> k = 0

=> x = y = z = 0

\(\Leftrightarrow\frac{x^2}{2}-\frac{x^2}{5}+\frac{y^2}{3}-\frac{y^2}{5}+\frac{z^2}{4}-\frac{z^2}{5}=0.\)

\(\Leftrightarrow\frac{3}{10}x^2+\frac{2}{15}y^2+\frac{1}{20}z^2=0\)(1)

Vế phải của (1) >= 0 với mọi x;y;z. Dấu "=" chỉ xảy ra khi x=y=z=0

Vậy nghiệm của đề bài là x=y=z=0.

\(\frac{6x^2+4y^2+3z^2}{12}=\frac{x^2+y^2+z^2}{5}\)

\(30x^2+20y^2+15z^2=12x^2+12y^2+12z^2\)

\(18x^2+8y^2+3z^2=0\)

\(18x^2\ge0\)   \(8y^2\ge0\)   \(3z^2\ge0\)

Nên

\(18x^2+8y^2+3z^2\ge0\)

Vậy

\(18x^2+8y^2+3z^2=0\) Khi và chỉ khi   \(18x^2=0\)  \(8y^2=0\)   \(3z^2=0\)

Vậy \(x=y=z=0\)