Question

[BiBo MoMo]

Cho m.n.p khác 0 và m+n-p/p= n+p-m/m=p+m-n/n

Tính H = (1+n/m)(1+p/n)(1+m/p)

Answer 1

[Nguyệt]

TH1: m+n+p khác 0

\(\frac{m+n-p}{p}=\frac{n+p-m}{m}=\frac{p+m-n}{n}\)

\(\Rightarrow2+\frac{m+n-p}{p}=2+\frac{n+p-m}{m}=2+\frac{p+m-n}{n}\)

\(\Rightarrow\frac{m+n+p}{p}=\frac{n+p+m}{m}=\frac{p+m+n}{n}\)

\(\Rightarrow p=m=n\)

thay m=n=p vào biểu thức H ta có:

\(H=\left(1+\frac{m}{m}\right).\left(1+\frac{n}{n}\right).\left(1+\frac{p}{p}\right)\)

\(H=2.2.2=2^3=8\)

TH2: m+n+p = 0 (m,n,p khác 0)

=> m=-(n+p)

=> n=-(m+p)

=>p=-(n+m)

thay m=-(n+p), n=-(m+p), p=-(n+m) vào biểu thức H

\(H=\left(1+\frac{-m-p}{m}\right).\left(1+\frac{-n-m}{n}\right).\left(1+\frac{-n-p}{p}\right)\)

\(H=\left(-\frac{p}{m}\right).\left(-\frac{m}{n}\right).\left(\frac{-n}{p}\right)=-1\)