Question

Cho a/x=b/x+1=c/x+2.Cmr:4(a-b)(b-c)=(a-c)^2

Answer 1

Áp dụng tính chất của dãy TSBN ta có:

+) \(\frac{a}{x}=\frac{b}{x+1}=\frac{a-b}{x-\left(x+1\right)}=\frac{a-b}{-1}\)

+) \(\frac{b}{x+1}=\frac{c}{x+2}=\frac{b-c}{\left(x+1\right)-\left(x+2\right)}=\frac{b-c}{-1}\)

+) \(\frac{a}{x}=\frac{c}{x+2}=\frac{a-c}{x-\left(x+2\right)}=\frac{a-c}{-2}\)

⇒ \(\left(\frac{a-b}{-1}\right)^2=\left(\frac{b-c}{-1}\right)^2=\left(\frac{a-c}{-2}\right)^2\)

⇒ \(\frac{\left(a-b\right)\left(b-c\right)}{1}=\frac{\left(a-c\right)^2}{4}\)

⇒ \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)