Câu 3:
$y'=\dfrac{(x+1)'}{2\sqrt{x+1}}=\dfrac{1}{2\sqrt{x+1}}$
$y''=\dfrac{-2(\sqrt{x+1})'}{4(x+1)}$
$=\dfrac{-2.\dfrac{1}{2\sqrt{x+1}}}{4x+4}$
$=\dfrac{-1}{\sqrt{x+1}(4x+4)}$
Ta có:
$4y^3y''+1$
$=4.(\sqrt{x+1})^3.\dfrac{-1}{\sqrt{x+1}(4x+4)}+1$
$=-(\sqrt{x+1})^2.\dfrac{1}{x+1}+1$
$=-1+1=0$ (đpcm)