Đáp án:
Năm 2012-2013
$a)$
$P=(\dfrac{4a}{√a-1}-$ $\dfrac{√a}{a-√a}).$ $(\dfrac{√a-1}{a^2})$
$=[\dfrac{4a}{√a-1}-$ $\dfrac{√a}{√a(√a-1)}].$ $\dfrac{√a-1}{a^2}$
$=\dfrac{4a√a-√a}{√a(√a-1)}.$ $\dfrac{√a-1}{a^2}$
$=\dfrac{√a(4a-1)}{√a(√a-1)}.$ $\dfrac{√a-1}{a^2}$
$=\dfrac{(4a-1)(√a-1)}{(√a-1)a^2}$
$=\dfrac{4a-1}{a^2}$ (Điều kiện: $a>0$ và $a$$\neq$$1$)
$b)$ Để $P=3$, thì:
$\dfrac{4a-1}{a^2}=3$
$⇔4a-1=3a^2$
$⇔3a^2-4a+1=0$
$⇔3a^2-3a-a+1=0$
$⇔3a(a-1)-(a-1)=0$
$⇔(a-1)(3a-1)=0$
$⇔a-1=0$ hoặc $3a-1=0$
$⇔ a=1 (loại)$ hoặc $a=1/3 (nhận)$
Vậy để $P=3 $ thì $a=1/3$
Năm 2013-2014$
$P=√12-√27-2√48$
$=\sqrt[]{2^2.3}-$ $\sqrt[]{3^2.3}-$ $\sqrt[]{2^2.48}$
$=2\sqrt[]{3}-$ $3\sqrt[]{3}-$ $\sqrt[]{192}$
$=2\sqrt[]{3}-$ $3\sqrt[]{3}-$ $\sqrt[]{8^2.3}$
$=2\sqrt[]{3}-$ $3\sqrt[]{3}-$ $8\sqrt[]{3}$
$=(2-3-8)\sqrt[]{3}$ $=-9\sqrt[]{3}$
$Q=(\dfrac{1}{√x+3}+$ $\dfrac{1}{√x-3}).$ $\dfrac{√x+3}{√x}$
$=\dfrac{√x-3+√x+3}{(√x+3)(√x-3)}.$ $\dfrac{√x+3}{√x}$
$=\dfrac{2√x(√x+3)}{√x(√x+3)(√x-3)}$
$=\dfrac{2}{√x-3}$ (Điều kiện: $x>0$; $x$$\neq$$9$)
BẠN THAM KHẢO NHA!!!
