Ta có: \(Q=\dfrac{x^2+x+1}{x^2+2x+1}\)
\(\Rightarrow\dfrac{1}{Q}=\dfrac{x^2+2x+1}{x^2+x+1}\)
Để Q min thì \(\dfrac{1}{Q}\) max
\(\dfrac{1}{Q}=\dfrac{x^2+2x+1}{x^2+x+1}=1+\dfrac{x}{x^2+x+1}\)
\(=1+\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{-x^2+2x+1}{x^2+x+1}=\dfrac{4}{3}-\dfrac{1}{3}.\dfrac{\left(-x-1\right)^2}{x^2+x+1}\le\dfrac{4}{3}\)
( Vì mẫu > 0 và tử \(\ge0\) )
\(\Rightarrow\dfrac{1}{Q}\) đạt GTNN là \(\dfrac{4}{3}\) khi x =1
Vậy Q đạt GTNN là \(\dfrac{3}{4}\) khi x = 1
Ta có: \(\dfrac{a+b}{a}=\dfrac{a}{b}\)
\(\Leftrightarrow\dfrac{a}{b}-1-\dfrac{1}{\dfrac{a}{b}}=0\)
\(\Leftrightarrow\left(\dfrac{a}{b}\right)^2-\dfrac{a}{b}-1=0\)
\(\Leftrightarrow\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{\sqrt{5}+1}{2}\\\dfrac{a}{b}=\dfrac{-\sqrt{5}+1}{2}\end{matrix}\right.\)
Thế \(\dfrac{a}{b}\) vào PT \(x^2-x-1\)
\(\Rightarrowđpcm\)
