Question

Cho \(a+b+c=1\) \(\left(1\right)\) ; \(a^2+b^2+c^2=1\) \(\left(2\right)\) ; \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\left(3\right)\)

CMR : \(xy+yz+zx=0\)

Answer 1

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\) thì \(x=ak;y=bk;z=ck\)

Khi đó \(xy+yz+zx=abk^2+ack^2+bck^2=k^2\left(ab+bc+ac\right)\left(4\right)\)

Từ \(\left(1\right)\) ta có : \(\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}=\frac{2}{\left(n-1\right)n\left(n+1\right)}\)

hay \(a^2+b^2+c^2+2ab+2bc+2ac=1.\)

Do \(\left(2\right)\) nên \(2ab+2ac+2bc=0\) tức là \(ab+bc+ac=0\)

Thay vào \(\left(4\right)\) được \(xy+yz+zx=0\).