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 .
Đặt \(\frac{x}{7}=\frac{y}{8}=\frac{z}{9}=k\Rightarrow x=7k;y=8k;z=9k\)
Thay x=7k;y=8k;z=9k vào A ta được:
\(A=\left(7k-8k\right)\left(8k-9k\right)-\left(\frac{7k-9k}{2}\right)^2\)=(-k)(-k)-\(\left(\frac{-2k}{2}\right)^2\)=k2-(-k)2=k2-k2=0
Đặt \(\frac{x}{7}=\frac{y}{8}=\frac{z}{9}=k\Rightarrow x=7k,y=8k,z=9k.\)
Thay vào biểu thức đã cho ta có
\(A=\left(7k-8k\right)\left(8k-9k\right)\left(\frac{7k-9k}{2}\right)^2=k^2-k^2=0.\)