Question

cho các số x,y,z khác 0 va thoả mãn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0.t\text{ính}gi\text{á}tr\text{ị}bi\text{ểu}th\text{ức}P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

Answer 1

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\)

\(P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

\(=\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}\)

\(=y\left(\frac{1}{x}+\frac{1}{z}\right)+x\left(\frac{1}{z}+\frac{1}{y}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(=y.\frac{-1}{y}+x.\frac{-1}{x}+z.\frac{-1}{z}\)

\(=-1-1-1=-3\)

Answer 2

P+3=\(\frac{y+z}{x}+1+\frac{x+z}{y}+1+\frac{x+y}{z}+1=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{x}\)

P+3=\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0.\left(x+y+z\right)=0\)

=> P=\(-3\)

Chuc ban hoc tot

Answer 3

Ta có : \(P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)

\(\Rightarrow P+3=\frac{y+z}{x}+1+\frac{z+x}{y}+1+\frac{x+y}{z}+1\)

\(\Rightarrow P+3=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}\)

\(\Rightarrow P+3=\left(x+y+z\right).\frac{1}{x}+\left(x+y+z\right).\frac{1}{y}+\left(x+y+z\right).\frac{1}{z}\)

\(\Rightarrow P+3=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\Rightarrow P+3=\left(x+y+z\right).0\)

\(\Rightarrow P+3=0\)

\(\Rightarrow P=-3\)

Vậy P = - 3