Đáp án:
a) $P=-\dfrac{\sqrt x}{x+\sqrt x+1}$
b) $P=-\dfrac{2}{7}$
Giải thích các bước giải:
Bài 12:
ĐKXĐ: $x\ge 0;x\ne 1$
a) $P=\dfrac{\sqrt x+1}{x-1}-\dfrac{x+2}{x\sqrt x-1}-\dfrac{\sqrt x+1}{x+\sqrt x+1}$
$=\dfrac{\sqrt x+1}{(\sqrt x+1)(\sqrt x-1)}-\dfrac{x+2}{\sqrt x^3-1}-\dfrac{(\sqrt x+1)(\sqrt x-1)}{(\sqrt x-1)(x+\sqrt x+1)}$
$=\dfrac{1}{\sqrt x-1}-\dfrac{x+2}{(\sqrt x-1)(x+\sqrt x+1)}-\dfrac{(\sqrt x+1)(\sqrt x-1)}{(\sqrt x-1)(x+\sqrt x+1)}$
$=\dfrac{x+\sqrt x+1-x-2-x+1}{(\sqrt x-1)(x+\sqrt x+1)}$
$=\dfrac{-\sqrt x(\sqrt x -1)}{(\sqrt x-1)(x+\sqrt x+1)}$
$=-\dfrac{\sqrt x}{x+\sqrt x+1}$
b) $x=\sqrt{7+4\sqrt 3}+\sqrt{7-4\sqrt 3}$
$=\sqrt{4+2.2.\sqrt 3+3}+\sqrt{4-2.2.\sqrt 3+3}$
$=\sqrt{(2+\sqrt 3)^2}+\sqrt{(2-\sqrt 3)^2}$
$=|2+\sqrt 3|+2-\sqrt 3|$
$=2+\sqrt 3+2-\sqrt 3$
$=4$
$⇒P=-\dfrac{\sqrt x}{x+\sqrt x+1}=-\dfrac{\sqrt 4}{4+\sqrt 4+1}=-\dfrac{2}{7}$
Vậy khi $x=\sqrt{7+4\sqrt 3}+\sqrt{7-4\sqrt 3}$ thì $P=-\dfrac{2}{7}$.
