=> \(\frac{ay+bx}{xy}=\frac{bz+cy}{yz}=\frac{cx+az}{zc}\) <=> \(\frac{a}{x}+\frac{b}{y}=\frac{b}{y}+\frac{c}{z}=\frac{c}{z}+\frac{a}{c}\) 

<=> \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=k\)=> \(x=ak\) ; \(y=bk\) ; \(z=ck\) (2)

Gọi giả thiết là (1)  Thay 2 vào 1 ta đc : \(k=\frac{1}{2}\)

=> Kết hợp k=1/2 với 2 ta được: a=x/2 ; b=y/2 và c=z/2

bạn lầu trên ơi, a/x=b/y=c/x=k thì x=a/k chứ bạn đâu phải x=ak đâu.

a) \(\frac{-2}{5}+\frac{5}{6}.x=\frac{-4}{15}\)

\(\frac{5}{6}.x=\frac{-4}{15}-\frac{-2}{5}\)

\(\frac{5}{6}.x=\frac{2}{15}\)

\(x=\frac{2}{15}:\frac{5}{6}\)

\(x=\frac{4}{25}\)

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

\(x-\frac{1}{5}=0\)

\(x=0+\frac{1}{5}\)

\(x=\frac{1}{5}\)

Ta có

\(\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|y+\frac{3}{2}\right|\ge0\\\left|x+y-z-\frac{1}{2}\right|\ge0\end{cases}\)

Maf \(\left|x-\frac{1}{2}\right|+\left|y+\frac{3}{2}\right|+\left|x+y-z-\frac{1}{2}\right|=0\)

\(\Rightarrow\begin{cases}x-\frac{1}{2}=0\\y+\frac{3}{2}=0\\x+y-z-\frac{1}{2}=0\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\x+y-z=\frac{1}{2}\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\\frac{1}{2}-\frac{3}{2}-z=\frac{1}{2}\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\-z=\frac{3}{2}\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\z=-\frac{3}{2}\end{cases}\)

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

đặt \(\frac{x}{7}=\frac{y}{8}=\frac{z}{9}=k\Rightarrow x=7k;y=8k;z=9k\)

=>A=\(\left(7k-8k\right)\left(8k-9k\right)-\left(\frac{7k-9k}{2}\right)^2=\left(-k\right)\left(-k\right)-\left(\frac{2k}{2}\right)^2\)

=k²-k²=0

Đặt \(\frac{x}{7}=\frac{y}{8}=\frac{z}{9}=k\)

\(\Rightarrow\hept{\begin{cases}x=7k\\y=8k\\z=9k\end{cases}}\left(1\right)\)

Thay (1) vào: \(A=\left(7k-8k\right)\left(8k-9k\right)-\left(\frac{7k-9k}{2}\right)^2\)

\(=-k.\left(-k\right)-\left(-k\right)^2\)

\(=k^2-k^2=0\)

Vậy A =0 .

b)  \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)

<=>  \(x^2+y^2+z^2+3-2x-2y-2x\ge0\)

<=>  \(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\ge0\)luôn đúng

Dấu "=" xảy ra  <=>  x=y=z=1