a,b,c khác 0 nhé

Ba số x,y,z tỉ lệ với ba số a,b,c

\(\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}\)(1)

Lại có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{xa}{a^2}+\frac{yb}{b^2}+\frac{zc}{c^2}=\frac{xa+yb+zc}{a^2+b^2+c^2}=\frac{9\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=9\) (2)

Từ (1) và (2) ta có : \(\frac{x+y+z}{a+b+c}=9\)

\(\Rightarrow\left(x+y+z\right)=9\left(a+b+c\right)\) (đpcm)

Theo bài ra ta có:\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)

\(\Rightarrow x=ak;y=bk;z=ck\)

Khi đó:\(ax+by+cz=a\cdot ak+b\cdot bk+c\cdot ck=a^2k+b^2k+c^2k=k\left(a^2+b^2+c^2\right)\)

Vì \(ax+by+cz=9\left(a^2+b^2+c^2\right)\Rightarrow k\left(a^2+b^2+c^2\right)=9\left(a^2+b^2+c^2\right)\)

\(\Rightarrow k=9\)

Khi đó:\(x+y+z=ak+bk+ck=k\left(a+b+c\right)=9\left(a+b+c\right)\left(đpcm\right)\)

Ta có: 2x=4y=3z

\(\frac{a+b-c}{6}=\frac{b+c-a}{10}=\frac{c+a-b}{2}=\frac{a}{4}=\frac{b}{8}=\frac{c}{6}\)

\(\Rightarrow\frac{2ãx}{4}=\frac{4by}{8}=\frac{3cz}{6}=\frac{ax}{2}=\frac{by}{2}=\frac{cz}{2}\)

\(\Rightarrowãx=by=cz\)

dòng thứ hai mk ko hiểu !

\(\frac{a+b-c}{6}=\frac{b+c-a}{10}=\frac{a+b-c+b+c-a}{6+10}=\frac{2b}{16}=\frac{b}{8}\)

kiểu như vậy đó Mk làm tắt sory sory

\(\left\{{}\begin{matrix}x^2-yz=a\\y^2-xz=b\\z^2-xy=c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^3-xyz=ax\\y^3-xyz=by\\z^3-xyz=cz\end{matrix}\right.\) \(\Rightarrow ax+by+cz=x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)⋮\left(x+y+z\right)\)