Question

Cho a>b>0.So sánh x,y biết

\(x=\frac{a+1}{1+a+a^2}\)

\(y=\frac{b+1}{1+b+b^2}\)

Answer 1

Ta có: \(a>b>0\)

   \(\Rightarrow a^2>b^2\)

\(\Rightarrow a^2+a>b^2+b\)

\(\Rightarrow a^2+a+1>b^2+b+1\)

\(\Rightarrow\frac{1}{a^2+a+1}< \frac{1}{b^2+b+1}\)

\(\Rightarrow x< y\)

Answer 2

\(x=\frac{a+1}{a^2+a+1}=1-\frac{a^2}{a+a+1}\)

\(y=\frac{b+1}{1+b+b^2}=1-\frac{b^2}{1+b+b^2}\)

Do \(\frac{a^2}{a^2+a+1}>\frac{b^2}{b^2+b+1}\Rightarrow x< y\)