Question

Answer 1

a: ĐKXD: a<>1

b: \(P=\left(\dfrac{a-1}{a^2+a+1}-\dfrac{a^2-3a+1}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{1}{a-1}\right)\cdot\dfrac{1-a}{a^2+1}\)

\(=\dfrac{a^2-2a+1-a^2+3a-1-a^2-a-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{1-a}{a^2+1}\)

\(=\dfrac{-a^2-1}{a^2+a+1}\cdot\dfrac{-1}{a^2+1}=\dfrac{1}{a^2+a+1}\)

 

Answer 2

c. \(\dfrac{1}{P}=\dfrac{1}{\dfrac{1}{x^2+x+1}}=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy \(MinP=\dfrac{3}{4}\)