Đáp án: $ y\ge \dfrac{13}{3}$
Giải thích các bước giải:
Ta có:
$y=\dfrac{2x^2+5}{x+1}$
$\to y=\dfrac{(2x^2-2)+7}{x+1}$
$\to y=\dfrac{2(x^2-1)+7}{x+1}$
$\to y=\dfrac{2(x-1)(x+1)+7}{x+1}$
$\to y=2(x-1)+\dfrac7{x+1}$
$\to y=2(x+1)+\dfrac7{x+1}-4$
$\to y=(\dfrac{7(x+1)}{9}+\dfrac{7}{x+1})+\dfrac{11}{9}(x+1)-4$
$\to y\ge 2\sqrt{\dfrac{7(x+1)}{9}\cdot \dfrac{7}{x+1}}+\dfrac{11}{9}\cdot (2+1)-4$
$\to y\ge \dfrac{13}{3}$
Dấu = xảy ra khi $x=2$
