Câu hỏi của Lê Minh Khuê

Question

Giả sử x+y+z=2017 và 1/x+y +1/y+z +1/x+z= 1/672
Tính tổng C = x/y+z + y/z+x + z/x+y

Nguyễn Đức Trí · Accepted Answer

$\left(x+y+z\right).\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{2017}{672}$

$\Rightarrow\left(\dfrac{x+y+z}{x+y}+\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{x+z}\right)=\dfrac{2017}{672}$

$\Rightarrow1+\dfrac{z}{x+y}+1+\dfrac{x}{y+z}+1+\dfrac{y}{x+z}=\dfrac{2017}{672}$

$\Rightarrow3+\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{2017}{672}$

$\Rightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{2017}{672}-3=\dfrac{2017}{672}-\dfrac{2016}{672}=\dfrac{1}{672}$

$\Rightarrow C=\dfrac{1}{672}$Câu trả lời: $\left(x+y+z\right).\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\right)=\dfrac{2017}{672}$

$\Rightarrow\left(\dfrac{x+y+z}{x+y}+\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{x+z}\right)=\dfrac{2017}{672}$

$\Rightarrow1+\dfrac{z}{x+y}+1+\dfrac{x}{y+z}+1+\dfrac{y}{x+z}=\dfrac{2017}{672}$

$\Rightarrow3+\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{2017}{672}$

$\Rightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}=\dfrac{2017}{672}-3=\dfrac{2017}{672}-\dfrac{2016}{672}=\dfrac{1}{672}$

$\Rightarrow C=\dfrac{1}{672}$