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

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

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

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

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

\(\Rightarrow C=\dfrac{1}{672}\)

x+y=1/2 :y+z=1/3 vay x-z=1/6

                                 z+x=1/4 vay2x=5/12 suy ra x=5/24 vay y=1/2-5/24=7/24

                                                                                    z=1/3-7/24=1/24

Từ đầu bài suy ra:

\(\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)

\(\Leftrightarrow x+y+y+z+z+x=\frac{13}{12}\)

\(\Leftrightarrow2x+2y+2z=\frac{13}{12}\)

\(\Leftrightarrow2\left(x+y+z\right)=\frac{13}{12}\)

\(\Rightarrow x+y+z=\frac{13}{12}:2=\frac{13}{24}\)

\(\Rightarrow x=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\)

\(y=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\)

\(z=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\)

Vậy...

x+y=1/2;y+z=1/3;z+x=1/4

=>2.(x+y+z)=1/2+1/3+1/4=13/12

x+y=1/2=>z=13/12-1/2=7/12

y+z=1/3=>x=13/12-1/3=3/4

z+x=1/4=>y=13/12-1/4=5/6

tim x,y,z biet y^2=y-1; x^2=x-1; z^2=z-1

NhOk ChỈ Là 1 FaN CuỒnG CủA KhẢi tra loi vay thi chet ho cai.

Ta có :

\(x+y=\frac{1}{2};y+z=\frac{1}{3};z+x=\frac{1}{6}\)

\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow2x+2y+2z=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\)

\(\Rightarrow2\left(x+y+z\right)=1\)

\(\Rightarrow x+y+z=\frac{1}{2}\)

\(\Rightarrow\hept{\begin{cases}\left(x+y+z\right)-\left(x+y\right)=\frac{1}{2}-\frac{1}{2}\Rightarrow z=0\\\left(x+y+z\right)-\left(y+z\right)=\frac{1}{2}-\frac{1}{3}\Rightarrow x=\frac{1}{6}\\\left(x+y+z\right)-\left(z+x\right)=\frac{1}{2}-\frac{1}{6}\Rightarrow y=\frac{1}{3}\end{cases}}\)

Vậy \(x=\frac{1}{6},y=\frac{1}{3};z=0\) .

\(x+y=\frac{1}{2};y+z=\frac{1}{3};z+x=\frac{1}{6}\)

Ta có:\(\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

\(\Leftrightarrow2\left(x+y+z\right)=1\)

\(\Leftrightarrow x+y+z=\frac{1}{2}\)

\(\Rightarrow\hept{\begin{cases}\left(x+y+z\right)-\left(x+y\right)=\frac{1}{2}-\frac{1}{2}=0\\\left(x+y+z\right)-\left(y+z\right)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\\\left(x+y+z\right)-\left(z+x\right)=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\end{cases}}\)

Vậy....

Đây nhé!

\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{7}{10}\)

Cộng thêm 3 vào mỗi vế ta được:

\(\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{z+x}+1\right)+\left(\frac{z}{x+y}+1\right)=\frac{7}{10}+3=\frac{37}{10}\)

Quy đồng mỗi cái biểu thức trong ngoặc lên,ta được:

\(\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}=\frac{37}{10}\)

Đặt thừa số chung ở biểu thức vế trái,ta được:

\(\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{37}{10}\)

Thay giả thiết đề bài vào,ta lại có:

\(\left(x+y+z\right).\frac{2}{5}=\frac{37}{10}\Rightarrow x+y+z=\frac{37}{10}:\frac{2}{5}=\frac{37}{4}\)

:D?