Đáp án:
Giải thích các bước giải:
$ĐKXĐ:x>0;x\neq4$
$a,A=\left (\dfrac{x-\sqrt{x}+7}{x-4}+\dfrac{1}{\sqrt{x}-2} \right ):\left (\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{2\sqrt{x}}{x-4} \right )$
$A=\left (\dfrac{x-\sqrt{x}+7+\sqrt{x}+2}{x-4} \right ):\left (\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{x-4} \right )$
$A=\left (\dfrac{x+9}{x-4} \right ).\left (\dfrac{x-4}{6\sqrt{x}} \right )$
$A=\dfrac{x+9}{6\sqrt{x}}$
$b,\dfrac{1}{A}=\dfrac{6\sqrt{x}}{x+9}$
$A-\dfrac{1}{A}=\dfrac{x+9}{6\sqrt{x}}-\dfrac{6\sqrt{x}}{x+9}$
$=\dfrac{(x+9)^2}{6\sqrt{x}(x+9)}-\dfrac{36x}{6\sqrt{x}(x+9)}$
$=\dfrac{x^2+18x+81-36x}{6\sqrt{x}(x+9)}$
$=\dfrac{x^2-18x+81}{6\sqrt{x}(x+9)}$
$=\dfrac{(x-9)^2}{6\sqrt{x}(x+9)}$
$\text{Vì $\dfrac{(x-9)^2}{6\sqrt{x}(x+9)}>0∀x$}$
$=>A>\dfrac{1}{A}$
Chúc bạn học tốt.
