cho x/y+z + y/z+x +z/x+y =1. Tính S=x^2/y+z +y^2/z+x + z^2/x+y 08/11/2021 Bởi Melanie cho x/y+z + y/z+x +z/x+y =1. Tính S=x^2/y+z +y^2/z+x + z^2/x+y
ta có x/y+z + y/x+z + z/x+y =1 => (x+y+z)(x/y+z + y/x+z + z/x+y) = x+y+z => x²+x(y+z)/y+z +y²+y(x+z)/x+z + z²+z(x+y)/x+y = x+y+z => x²/y+z + x+ y²/x+z +y+ z²/x+y + z = x+ y+z => x²/y+z + y²/x+z + z²/x+y = 0 vậy S = 0 Bình luận
Ta có: `x/(y+z)+y/(x+z)+z/(x+y)=1` `⇒(x+y+z)(x/(y+z)+y/(x+z)+z/(x+y))=x+y+z` `⇒(x^2+x(y+z))/(y+z)+(y^2+y(x+z))/(x+z)+(z^2+z(x+y))/(x+y)=x+y+z` `⇒x^2/(y+z)+x+y^2/(z+x)+y+z^2/(x+y)+z=x+y+z` `⇒x^2/(y+z)+y^2/(z+x)+z^2/(x+y)=0` Vậy `S=0` Bình luận
ta có x/y+z + y/x+z + z/x+y =1
=> (x+y+z)(x/y+z + y/x+z + z/x+y) = x+y+z
=> x²+x(y+z)/y+z +y²+y(x+z)/x+z + z²+z(x+y)/x+y = x+y+z
=> x²/y+z + x+ y²/x+z +y+ z²/x+y + z = x+ y+z
=> x²/y+z + y²/x+z + z²/x+y = 0
vậy S = 0
Ta có: `x/(y+z)+y/(x+z)+z/(x+y)=1`
`⇒(x+y+z)(x/(y+z)+y/(x+z)+z/(x+y))=x+y+z`
`⇒(x^2+x(y+z))/(y+z)+(y^2+y(x+z))/(x+z)+(z^2+z(x+y))/(x+y)=x+y+z`
`⇒x^2/(y+z)+x+y^2/(z+x)+y+z^2/(x+y)+z=x+y+z`
`⇒x^2/(y+z)+y^2/(z+x)+z^2/(x+y)=0`
Vậy `S=0`