a: \(P=\dfrac{a+1+\sqrt{a}}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)

\(=\dfrac{a+\sqrt{a}+1}{a+1}\cdot\dfrac{\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}=\dfrac{a+\sqrt{a}+1}{\sqrt{a}-1}\)

b: P<1

=>P-1<0

=>\(\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)

=>căn a-1<0

=>0<a<1

c: Thay x=19-8căn3 vào P, ta được:

\(P=\dfrac{19-8\sqrt{3}+4+\sqrt{3}+1}{4+\sqrt{3}-1}=\dfrac{31-15\sqrt{3}}{2}\)

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x-2\ne0\\x-2\ge0\end{matrix}\right.\Leftrightarrow x>2\\ b,ĐK:\dfrac{1}{3-2x}\ge0\Leftrightarrow3-2x\ge0\left(1>0\right)\Leftrightarrow x\le\dfrac{3}{2}\)

\(2,\\ a,=\sqrt{\left(6-\sqrt{35}\right)\left(6+\sqrt{35}\right)}=\sqrt{36-35}=\sqrt{1}=1\\ b,=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}=\sqrt{81-17}=\sqrt{64}=8\\ c,=4\sqrt{2}-6\sqrt{6}+9-4\sqrt{2}+6\sqrt{6}=9\\ d,=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-2-\sqrt{3}=\sqrt{3}+\sqrt{2}-2-\sqrt{3}=\sqrt{2}-2\\ e,=\left(200\sqrt{3}-225\sqrt{3}+25\sqrt{3}\right):\sqrt{15}=0:\sqrt{15}=0\)

= 101 - 2 = 99

Xét ΔABD vuông tại A có AH là đường cao

nên \(AH^2=HD\cdot HB\)

=>\(AH=\sqrt{20,25\cdot9}=13,5\left(cm\right)\)

Xét ΔDAC vuông tại D có DH là đường cao

nên \(DH^2=AH\cdot HC\)

=>\(HC=\dfrac{20.25^2}{13.5}=30,375\left(cm\right)\)

BD=BH+DH=9+20,25=29,25(cm)

AC=AH+HC=13,5+30,375=43,875(cm)

Vì AC\(\perp\)BD tại H

nên \(S_{ABCD}=\dfrac{1}{2}\cdot AC\cdot BD=\dfrac{1}{2}\cdot29,25\cdot43,875\simeq641,7\left(cm^2\right)\)

Lời giải:
a. Gọi biểu thức là $A$

 \(A=\left[\frac{(1-\sqrt{x})(1+\sqrt{x}+x)}{1-\sqrt{x}}+\sqrt{x}\right].\left[\frac{1-\sqrt{x}}{(1-\sqrt{x})(1+\sqrt{x})}\right]^2\)

\(=(1+\sqrt{x}+x+\sqrt{x}).\frac{1}{(1+\sqrt{x})^2}=(\sqrt{x}+1)^2.\frac{1}{(\sqrt{x}+1)^2}=1\)

Ta có đpcm

b.

\(\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}=\sqrt{2011}-\sqrt{2010}\)