Question

tìm GTLN

\(\dfrac{3-4x}{x^2+1}\)

Answer 1

\(=\dfrac{3-4x}{x^2+1}-4+4\)

\(=\dfrac{3-4x-4x^2-4}{x^2+1}+4\)

\(=\dfrac{-4x^2-4x-1}{x^2+1}+4\)

\(=\dfrac{-\left(2x+1\right)^2}{x^2+1}+4\le4\)

dấu = xảy ra khi 2x+1=0<=>x=\(\dfrac{-1}{2}\)

vậy GTLN=4 khi \(x=\dfrac{-1}{2}\)

Answer 2

\(=\dfrac{3-4x}{x^2+1}+1-1\)

\(=\dfrac{3-4x+x^2+1}{x^2+1}-1\)

\(=\dfrac{x^2-4x+4}{x^2+1}-1\)

\(=\dfrac{\left(x-2\right)^2}{x^2+1}-1\ge-1\)

dấu = xảy ra khi x-2=0<=>x=2

vậy GTNN=-1 khi x=2