cho : $\frac{1}{x}$ + $\frac{1}{y}$ + $\frac{1}{z}$ = $\frac{1}{x+y+z}$
chứng minh rằng : $\frac{1}{x mũ 2019}$ + $\frac{1}{y mũ 2019}$ + $\frac{1}{z mũ 2019}$ = $\frac{1}{x mũ 2019 + y mũ 2019 + z mũ 2019}$
cho : $\frac{1}{x}$ + $\frac{1}{y}$ + $\frac{1}{z}$ = $\frac{1}{x+y+z}$
chứng minh rằng : $\frac{1}{x mũ 2019}$ + $\frac{1}{y mũ 2019}$ + $\frac{1}{z mũ 2019}$ = $\frac{1}{x mũ 2019 + y mũ 2019 + z mũ 2019}$
Giải thích các bước giải :
`1/x+1/y+1/z=1/(x+y+z)`
`<=>(1/x+1/y)+(1/z-1/(x+y+z))=0`
`<=>(x+y)/(xy)+(x+y+z)/[z(x+y+z)]-z/[z(x+y+z)]=0`
`<=>(x+y)/(xy)+(x+y+z-z)/[z(x+y+z)]=0`
`<=>(x+y)/(xy)+(x+y)/[z(x+y+z)]=0`
`<=>(x+y)(1/(xy)+1/[z(x+y+z)])=0`
`<=>(x+y)([z(x+y+z)]/[xyz(x+y+z)]+(xy)/[xyz(x+y+z)])=0`
`<=>(x+y)((xz+yz+z^2)/[xyz(x+y+z)]+(xy)/[xyz(x+y+z)])=0`
`<=>(x+y)(xz+yz+z^2+xy)/[xyz(x+y+z)]=0`
`<=>(x+y)([x(y+z)+z(y+z)]/[xyz(x+y+z)])=0`
`<=>[(x+y)(y+z)(z+x)]/[xyz(x+y+z)]=0`
`<=>(x+y)(y+z)(z+x)=0`
*)Th1: `x+y=0`
`<=>x=-y`
`+)1/(x^(2019))+1/(y^(2019))+1/(z^(2019))`
`=1/[(-y)^(2019)]+1/(y^(2019))+1/(z^(2019))`
`=(y^(2019))/[(-y)^(2019).y^(2019)]+(-y)^(2019)/[(-y)^(2019).y^(2019)]+1/(z^(2019))`
`=1/(z^(2019))` (1)
`+)1/(x^(2019)+y^(2019)+z^(2019))`
`=1/[(-y)^(2019)+y^(2019)+z^(2019)]`
`=1/(z^(2019))` (2)
Từ `(1)` và `(2)`
`=>1/x^(2019)+1/(y^2019)+1/z^(2019)=1/(x^(2019)+y^(2019)+z^(2019))`
Bạn cm 2 Th còn lại tương tự ạ
~Chúc bạn học tốt !!!~