Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có ax + by = c ; by + cz = a
<=> cz - ax = a - c (1)
mà cz + ax = b (2)
Từ (1) và (2) => \(cz=\frac{a-c+b}{2}\Rightarrow z=\frac{a-c+b}{2c}\Rightarrow z+1=\frac{a+b+c}{2c}\)
=> \(\frac{1}{z+1}=\frac{2c}{a+b+c}\)
Tương tự ta có \(\frac{1}{x+1}=\frac{2a}{a+b+c}\); \(\frac{1}{y+1}=\frac{2b}{a+b+c}\)
=> P = \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}=2\)
\(2x-2y=by+cz-cz-ax=by-ax\)
\(\Rightarrow2x-2y=by-ax\)
\(\Rightarrow2x+ax=2y+by\)
\(\Rightarrow x\left(a+2\right)=y\left(b+2\right)\)
\(\Rightarrow a+2=\dfrac{y\left(b+2\right)}{x}\)
\(2z-2y=ax+by-cz-ax=by-cz\)
\(\Rightarrow2z+cz=2y+by\)
\(\Rightarrow z\left(c+2\right)=y\left(b+2\right)\)
\(\Rightarrow c+2=\dfrac{y\left(b+2\right)}{z}\)
\(A=\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}=\dfrac{2}{\dfrac{y\left(b+2\right)}{x}}+\dfrac{2}{b+2}+\dfrac{2}{\dfrac{y\left(b+2\right)}{z}}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2}{b+2}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2y}{y\left(b+2\right)}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x+2y+2z}{y\left(b+2\right)}=\dfrac{by+cz+cz+ax+ax+by}{by+2y}=\dfrac{2\left(ax+by+cz\right)}{by+cz+ax}=2\)
