#)Giải :

Đặt \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=k\Rightarrow\hept{\begin{cases}a=kx\\b=ky\\c=kz\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(a^2+2b^2+3c^2\right)\left(x^2+2y^2+3z^2\right)=\left[\left(kx\right)^2+2\left(ky\right)^2+3\left(kz\right)^2\right]\left(x^2+2y^2+3z^2\right)=k^2\left(a^2+2b^2+3c^2\right)^2\left(1\right)\\\left(ax+2by+3cz\right)^2=\left(kx.x+2ky.y+3kz.z\right)^2=\left[k\left(a^2+2b^2+3c^2\right)\right]^2=k^2\left(a^2+2b^2+3c^2\right)^2\left(2\right)\end{cases}}\)

Từ (1) và (2) => đpcm

Lời giải:

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=m\Rightarrow x=am; y=bm; z=cm\)

Khi đó:

\((x^2+2y^2+3z^2)(a^2+2b^2+3c^2)=[(am)^2+2(bm)^2+3(cm)^2](a^2+2b^2+3c^2)\)

\(=m^2(a^2+2b^2+3c^2)^2(1)\)

Và:

\((ax+2by+3cz)^2=(a.am+2b.bm+3c.cm)^2=[m(a^2+2b^2+3c^2)]^2\)

\(=m^2(a^2+2b^2+3c^2)^2(2)\)

Từ (1) và (2) ta có đpcm.

Đề sai rồi bạn

Vế phải : \(\left(ax+aby+3cz\right)^2\)

\(=a^2x^2+a^2b^2y^2+9c^2z^2+2a^2bxy+6abcyz+6axcz\)

Vế trái : \(\left(x^2+2y^2+3z^2\right)\left(a^2+2b^2+3c^2\right)\)

\(=\left(x^2+2y^2+3z^2\right)a^2+\left(x^2+2y^2+3z^2\right)2b^2+\left(x^2+2y^2+3z^2\right)3c^2\)

\(=x^2a^2+2a^2y^2+3a^2z^2+2b^2x^2+4b^2y^2+6b^2z^2+3x^2c^2+6c^2y^2+9c^2z^2\)

Ko hề bằng nhau 

\(\Rightarrow\)đề sai