Question

[Hoàng Nguyên Chiến]

Tìm Min hoặc Max của: N=x^2+x+1/x^2+1

Answer 1

[Lê Song Phương]

\(N=\dfrac{x^2+x+1}{x^2+1}\)

\(\Leftrightarrow\left(x^2+1\right)N=x^2+x+1\)

\(\Leftrightarrow Nx^2+N=x^2+x+1\)

\(\Leftrightarrow\left(N-1\right)x^2-x+N-1=0\) (*)

Có \(\Delta=\left(-1\right)^2-4\left(N-1\right)^2\) 

\(=1-\left(2N-2\right)^2\)

\(=\left(1+2N-2\right)\left(1-2N+2\right)\)

\(=\left(2N-1\right)\left(3-2N\right)\)

Để (*) có nghiệm thì \(\Delta=\left(2N-1\right)\left(3-2N\right)\ge0\)

\(\Leftrightarrow\dfrac{1}{2}\le N\le\dfrac{3}{2}\)

\(N=\dfrac{1}{2}\) khi \(x=-\dfrac{-1}{2\left(N-1\right)}=\dfrac{1}{2\left(\dfrac{1}{2}-1\right)}=-1\)

\(N=\dfrac{3}{2}\) khi \(x=-\dfrac{-1}{2\left(N-1\right)}=\dfrac{1}{2\left(\dfrac{3}{2}-1\right)}=1\)

Vậy \(minN=\dfrac{1}{2}\) khi \(x=-1\) và \(maxN=\dfrac{3}{2}\) khi \(x=1\)