Đáp án:
$\min A =\dfrac23 \Leftrightarrow x = y =\dfrac12$
Giải thích các bước giải:
Ta có:
$\quad x + y = 1$
$\Leftrightarrow y = 1 - x$
Ta được:
$\quad A = \dfrac{x}{y+1} +\dfrac{y}{x+1}$
$\to A =\dfrac{x}{2-x} +\dfrac{1-x}{x+1}$
$\to A =\dfrac{2(x^2-x+1)}{(2-x)(x+1)}$
$\to A =\dfrac{6x^2 - 6x + 6}{3(2-x)(x+1)}$
$\to A =\dfrac{2(4x^2 - 4x +1) - 2(x^2 - x - 2)}{3(2-x)(x+1)}$
$\to A =\dfrac{2(2x-1)^2}{3(2-x)(x+1)}+ \dfrac23$
$\to A \geq \dfrac23$
Dấu $=$ xảy ra $\Leftrightarrow 2x-1 = 0 \Leftrightarrow x =\dfrac12 \Rightarrow y =\dfrac12$
Vậy $\min A =\dfrac23 \Leftrightarrow x = y =\dfrac12$
