Toán 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 By Julia 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: $\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1$ $\to (x+y+z)(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y})=1.(x+y+z)$ $\to (x+y+x)(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y})=x+y+z$(*) Mà ta có: $(x+y+z)(\dfrac{x}{y+z}+\dfrac{x^2}{y+z}+\dfrac{xy}{y+z}+\dfrac{xz}{y+z}$ $\to = \dfrac{x^2}{y+z}+\dfrac{x(y+z)}{y+z}$(**) $\to = \dfrac{x^2}{y+z}+x$ Tương tự: $(x+y+z)\dfrac{y}{z+x}=\dfrac{y^2}{z+x}+y$(***) $(x+y+x)\dfrac{x}{x+y}=\dfrac{z^2}{x+y}+z$(****) Từ (*),(**),(***),(****),suy ra: $\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}=x+y+z$ $\to \dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0$ Mà: $S=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}$ $\to S=0$ Trả lời
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` Trả lời
Ta có:
$\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1$
$\to (x+y+z)(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y})=1.(x+y+z)$
$\to (x+y+x)(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y})=x+y+z$(*)
Mà ta có:
$(x+y+z)(\dfrac{x}{y+z}+\dfrac{x^2}{y+z}+\dfrac{xy}{y+z}+\dfrac{xz}{y+z}$
$\to = \dfrac{x^2}{y+z}+\dfrac{x(y+z)}{y+z}$(**)
$\to = \dfrac{x^2}{y+z}+x$
Tương tự:
$(x+y+z)\dfrac{y}{z+x}=\dfrac{y^2}{z+x}+y$(***)
$(x+y+x)\dfrac{x}{x+y}=\dfrac{z^2}{x+y}+z$(****)
Từ (*),(**),(***),(****),suy ra:
$\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}=x+y+z$
$\to \dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0$
Mà:
$S=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}$
$\to S=0$
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`