Question

Tìm x, y, z biết :
\(\dfrac{1+3y}{12}\) = \(\dfrac{1+5y}{5x}\)= \(\dfrac{1+7y}{4x}\)= \(\dfrac{z}{2}\)

Answer 1

Ta có:

\(\dfrac{1+3y}{12}=\dfrac{1+7y}{4x}=\dfrac{1+1+3y+7y}{12+4x}\)

\(=\dfrac{2+10y}{2.\left(6+2x\right)}=\dfrac{2.\left(1+5y\right)}{2.\left(6+2x\right)}=\dfrac{1+5y}{6+2x}=\dfrac{1+5y}{5x}\)

- Xét \(1+5y=0\Rightarrow y=\dfrac{-1}{5}\Rightarrow1+5y=0\) ( loại )

- Xét \(1+5y\ne0\Rightarrow6+2x=5x\)

\(\Rightarrow5x-2x=6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

Mà \(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}\)

\(\Rightarrow\dfrac{1+3y}{12}=\dfrac{1+5y}{10}\)

\(\Rightarrow10.\left(1+3y\right)=12.\left(1+5y\right)\)

\(\Rightarrow10+30y=12+60y\)

\(\Rightarrow10-12=60y-30y\)

\(\Rightarrow-2=30y\)

\(\Rightarrow y=\dfrac{-1}{5}\)

Vậy \(x=2\) , \(y=\dfrac{-1}{5}\)