Hướng dẫn giải:
Bài `18:`
`a,` `x=(-9)/(-25)=9/25=(9k)/(25k)(k\inZZ;k\ne0)`
`b,` `y=5/6=(5k)/(6k)(k\inZZ;k\ne0)`
`c,` `z=-0,36=(-9)/25=(-9k)/(25k)(k\inZZ;k\ne0)`
Bài `19:`
`8<2^n<2.32`
`<=>2^3<2^n<2.2^5`
`<=>2^3<2^n<2^6`
`<=>3<n<6`
`<=>n\in{4;5}`
Vậy `n\in{4;5}`
Bài `20:`
Vì `xyz=1` nên `x,y,z\ne0`
Ta có:
`@1/(1+x+xy)=z/(z(1+x+xy))=z/(z+xz+xyz)=z/(z+xz+1)`
`@1/(1+y+yz)=(xz)/(xz(1+y+yz))=(xz)/(xz+xzy+xzyz)=(xz)/(xz+1+z)`
Từ đó suy ra:
`P=1/(1+x+xy)+1/(1+y+yz)+1/(1+z+xz)`
`=z/(z+xz+1)+(xz)/(xz+1+z)+1/(1+z+xz)`
`=(z+xz+1)/(z+xz+1)`
`=1`
Vậy `P=1`
