Question

tìm x,y,z\(\varepsilon\) Q biết

x+y=7/6

y+z=1/14

x+z=1/12

 

Answer 1

Từ đầu bài suy ra:

\(\left(x+y\right)+\left(y+z\right)+\left(x+z\right)=\frac{7}{6}+\frac{1}{14}+\frac{1}{12}\)

\(\Rightarrow x+y+y+z+x+z=\frac{98}{84}+\frac{6}{84}+\frac{7}{84}\)

\(\Rightarrow2x+2y+2z=\frac{111}{84}\)

\(\Rightarrow2\left(x+y+z\right)=\frac{37}{28}\)

\(\Rightarrow x+y+z=\frac{37}{28}:2=\frac{37}{28}.\frac{1}{2}=\frac{37}{56}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{37}{56}-\frac{1}{14}=\frac{33}{56}\\y=\frac{37}{56}-\frac{1}{12}=\frac{97}{168}\\z=\frac{37}{56}-\frac{7}{6}=-\frac{185}{168}\end{cases}}\)

Vậy \(x=\frac{33}{56};y=\frac{97}{168};z=-\frac{185}{168}\)

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