Giải thích các bước giải:
Ta có :
$B=\dfrac{1}{3-\sqrt{x}}+\dfrac{\sqrt{x}}{3+\sqrt{x}}-\dfrac{x+9}{x-9}$
$\to B=\dfrac{3+\sqrt{x}}{(3+\sqrt{x})(3-\sqrt{x})}+\dfrac{\sqrt{x}(3-\sqrt{x})}{(3+\sqrt{x})(3-\sqrt{x})}+\dfrac{x+9}{9-x}$
$\to B=\dfrac{3+\sqrt{x}}{(3+\sqrt{x})(3-\sqrt{x})}+\dfrac{\sqrt{x}(3-\sqrt{x})}{(3+\sqrt{x})(3-\sqrt{x})}+\dfrac{x+9}{(3+\sqrt{x})(3-\sqrt{x})}$
$\to B=\dfrac{3+\sqrt{x}+\sqrt{x}(3-\sqrt{x})+x+9}{(3+\sqrt{x})(3-\sqrt{x})}$
$\to B=\dfrac{3+\sqrt{x}+3\sqrt{x}-x+x+9}{(3+\sqrt{x})(3-\sqrt{x})}$
$\to B=\dfrac{4\sqrt{x}+12}{(3+\sqrt{x})(3-\sqrt{x})}$
$\to B=\dfrac{4(\sqrt{x}+3)}{(3+\sqrt{x})(3-\sqrt{x})}$
$\to B=\dfrac{4}{3-\sqrt{x}}$
