Cho tỉ lệ thức x/y+z+1=y/x+z+1=z/x+y-z .Khi đó x+y/z+1=?
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Giải:
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=\frac{x+y+z}{y+z+1+x+z+1+x+y-2}=\frac{x+y+z}{2x+2y+2z}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)
\(\Rightarrow x+y+z=1\)
Xét \(\frac{x}{y+z+1}=\frac{1}{2}\)
\(\Rightarrow2x=y+z+1\)
\(\Rightarrow3x=x+y+z+1\)
\(\Rightarrow3x=1+1\)
\(\Rightarrow x=\frac{2}{3}\)
Xét \(\frac{y}{x+z+1}=\frac{1}{2}\)
\(\Rightarrow2y=x+z+1\)
\(\Rightarrow3y=x+y+z+1\)
\(\Rightarrow3y=1+1\)
\(\Rightarrow y=\frac{2}{3}\)
Xét \(\frac{z}{x+y-2}=\frac{1}{2}\)
\(\Rightarrow2z=x+y-2\)
\(\Rightarrow3z=x+y+z-2\)
\(\Rightarrow3z=1-2\)
\(\Rightarrow z=\frac{-1}{3}\)
Từ đó \(\frac{x+y}{z+1}=\frac{\frac{2}{3}+\frac{2}{3}}{\frac{-1}{3}+1}=\frac{\frac{4}{3}}{\frac{2}{3}}=\frac{4}{2}=2\)
Vậy \(\frac{x+y}{z+1}=2\)
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\(\frac{x}{y+z+1}\)= \(\frac{y}{x+z+1}\)= \(\frac{z}{x+y-2}\)= \(\frac{x+y+z}{y+z+1+x+z+1+x+y-1}\)
= \(\frac{x+y+z}{2x+2y+2z}\)= \(\frac{x+y+z}{2\left(x+y+z\right)}\)= \(\frac{1}{2}\)
=> \(\frac{z}{x+y-2}\)= \(\frac{1}{2}\)= \(\frac{z+1}{x+y-2+2}\)= \(\frac{z+1}{x+y}\)
=> \(\frac{z+1}{x+y}\)= \(\frac{1}{2}\)=> \(\frac{x+y}{z+1}\)= 2