Question

[Hoàng Lê Minh]

Giúp mình với mình cần gấp: cho mnp khác 0, am+bn+cp khác 0 thỏa mãn m=bn+cp,n=am+cp,p=am+bn tính giá trị biểu thức A=1/1+a + 1/1+b +1/1+c

Answer 1

[Nguyễn Linh Chi]

Ta có: \(m+n+p=2ma+2np+2pc\Rightarrow ma+np+pc=\frac{1}{2}\left(m+n+p\right)\)(1)

lại  có: 

\(\hept{\begin{cases}m=bn+cp\\n=am+cp\\p=am+bn\end{cases}\Rightarrow}\hept{\begin{cases}m-n=bn-am\\n-p=cp-bn\\p-m=am-cp\end{cases}}\Rightarrow\hept{\begin{cases}m\left(a+1\right)=n\left(b+1\right)\\n\left(b+1\right)=p\left(c+1\right)\\p\left(c+1\right)=m\left(a+1\right)\end{cases}}\)

\(\Rightarrow\frac{1}{m\left(a+1\right)}=\frac{1}{n\left(b+1\right)}=\frac{1}{p\left(c+1\right)}=\frac{3}{ma+mb+mc+m+n+p}\)( Dãy tỉ số bằng nhau)

\(=\frac{3}{\frac{1}{2}\left(m+n+p\right)+n+m+p}=\frac{2}{n+m+p}\)

=> \(\frac{1}{a+1}=\frac{2m}{m+n+p}\)

\(\frac{1}{b+1}=\frac{2n}{m+n+p}\)

\(\frac{1}{c+1}=\frac{2p}{m+n+p}\)

=> \(A=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2m+2n+2p}{m+n+p}=2\)