Question

cho ba so x,y,z thoa man : 19/x+y+19/y+z+19/z+x=7x/y+z+7y/z+x+7z/x+z=133/10

 

Answer 1

19(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}) = 7+\frac{7x}{y+z}+7+\frac{7y}{z+x}+7+\frac{7z}{z+y} - 21 \\ \\ 19(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}) = \frac{7(x+y+z)}{x+y}+\frac{7(x+y+z)}{y+z}+\frac{7(z+y+z)}{x+z} - 21 \\ \\ 19(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}) = 7(x+y+z).(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}) - 21 \\ \\ \frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z} = t \\ \\ \Rightarrow 19t = 7(x+y+z).t -21 = \frac{133}{10} \\ \\ 19t = \frac{133}{10} \Rightarrow t = \frac{7}{10} \\ \\ \Rightarrow 7(x+y+z).\frac{7}{10} -21 = \frac{133}{10} \Rightarrow M = x+y+z = 7

Answer 2

ban nguyen dan go the nao ra toan ki hieu la the