Tính tổng x + y + z biết $\frac{19}{x+y}$ + $\frac{19}{z+y}$ + $\frac{19}{x+z}$ = $\frac{7z}{x+y}$ + $\frac{7x}{z+y}$ + $\frac{7y}{x+z}$ = $\frac{1

Đáp án+Giải thích các bước giải:

 Ta có:

    `(19)/(x+y)+(19)/(z+y)+(19)/(x+z)=(133)/(10)`

`⇒19.(1/(x+y)+1/(z+y)+1/(x+z))=(133)/(10)`

`⇒1/(x+y)+1/(z+y)+1/(x+z)=(133)/(10) : 19`

`⇒1/(x+y)+1/(z+y)+1/(x+z)=7/(10)`

Ta có:

`(7z)/(x+y)+(7x)/(z+y)+(7y)/(x+z)=(133)/(10)`

`⇒7.(z/(x+y)+x/(z+y)+y/(x+z))=(133)/(10)`

`⇒z/(x+y)+x/(z+y)+y/(x+z)=(133)/(10) : 7`

`⇒z/(x+y)+x/(z+y)+y/(x+z)=19/10`

`⇒(z/(x+y)+1)+(x/(z+y)+1)+(y/(x+z)+1)=19/10+3`

`⇒(x+y+z)/(x+y)+(x+y+z)/(z+y)+(x+y+z)/(x+z)=19/10+30/10`

`⇒(x+y+z).(1/(x+y)+1/(z+y)+1/(x+z))=49/10`

`⇒(x+y+z). 7/(10)=49/10`

`⇒x+y+z=49/10 : 7/10`

`⇒x+y+z=7`

Vậy `x+y+z=7`