a, Khi x = 2, ta được: 

\(A=\dfrac{4}{2\sqrt{2}-2}=2+2\sqrt{2}\)

b, \(B=\dfrac{\sqrt{x}-4}{x-2\sqrt{x}}+\dfrac{3}{\sqrt{x}-2}\\ \Rightarrow B=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ \Rightarrow B=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(P=B:A=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{4}=-\left(\sqrt{x}-1\right)=1-\sqrt{x}\) (đpcm)

tick cho em la em lam lien

1. Với x = 36
=> A= \(\dfrac{\sqrt{36}-2}{\sqrt{36}-1}\)=\(\dfrac{4}{5}\)
2. Với x >0, x ≠1
B=\(\dfrac{x-5}{x-1}-\dfrac{2}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}-1}\)
B=\(\dfrac{x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{4\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
B=\(\dfrac{x-5-2\left(\sqrt{x}-1\right)+4\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
B=\(\dfrac{x-5-2\sqrt{x}+2+4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
B=\(\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
B=\(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
B=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
3. P=\(\dfrac{A}{B}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\). \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Ta có \(\sqrt{P}< \dfrac{1}{2}\)
=>P<\(\dfrac{1}{4}\)
=> \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)<\(\dfrac{1}{4}\)
=> \(4\left(\sqrt{x}-2\right)< \sqrt{x}+1\)
=> \(4\sqrt{x}-8< \sqrt{x}+1 \)
=> \(3\sqrt{x}< 9\)
=>\(\sqrt{x}< 3\)
=> x< 9
Lại có x ϵ Z => x ϵ {-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8}
Ta thử lại với x ≠ 1
=> x ϵ {-8,-7,-6,-5,-4,-3,-2,0,2,3,4,5,6,7,8}

Điều kiện: \(x\ge0,x\ne1\)

\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}+1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

Ta có \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0,\forall x\Rightarrow A>0\)

Lại có: \(A-2=\dfrac{2}{x+\sqrt{x}+1}-2=\dfrac{-2\left(x+\sqrt{x}\right)}{x+\sqrt{x}+1}\)

Mà \(x+\sqrt{x}+1>0;x+\sqrt{x}>0\) với mọi \(x\in TXĐ\)

\(\Rightarrow A-2< 0\Rightarrow A< 2\)

Vậy \(0< A< 2\)

a, Ta có : \(x=9\Rightarrow\sqrt{x}=3\)

Thay vào biểu thức A ta được : \(A=\frac{2}{3-2}=2\)

b, Với \(x\ge0;x\ne4\)

\(B=\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{4\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)+4\sqrt{x}}{x-4}\)

\(=\frac{x+2\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}=\frac{\sqrt{x}}{\sqrt{x}-2}\)( đpcm )

c, Ta có : \(A+B=\frac{3x}{\sqrt{x}-2}\)hay 

\(\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{2+\sqrt{x}}{\sqrt{x}-2}=\frac{3x}{\sqrt{x}-2}\)

\(\Rightarrow2+\sqrt{x}=3x\Leftrightarrow3x-2-\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}=3x-2\Leftrightarrow x=9x^2-12x+4\)

\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\Leftrightarrow x=\frac{4}{9}\left(ktm\right);x=1\)( đk : \(x\ge\frac{2}{3}\))

sao cô cho cả đáp án ra lun thế ạ @@

à ko em nhầm nhầm em xin lỗi cô 

1) A = 2 khi x =9

3) 2 + \(\sqrt{x}\)- 3x = 0

-3x +\(\sqrt{x}\)+2 = 0

3x- \(\sqrt{x}\)- 2 =0

Đặt \(\sqrt{x}\)= t ; ĐK : t\(\ge\)0; t \(\ne\) 2

\(^{3t^2-t-2=0}\)

t1 =1 (TMĐK)

t2 = \(\dfrac{-2}{3}\)(Không TMĐK)

Với t = 1 => \(\sqrt{x}=1\Leftrightarrow x=1\)

a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)

Thay vào biểu thức A ta được : \(\frac{1}{2-1}=1\)

b, Với \(x\ge0;x\ne1\)

\(Q=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{x-1}-1=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-2-x+1}{x-1}\)

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

c, Ta có : \(\frac{1}{Q}+P\le4\)hay\(1:\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\le4\)ĐK : \(x\ne1\)

\(\Leftrightarrow\frac{x-1+1}{\sqrt{x}-1}-4\le0\Leftrightarrow\frac{x-4\sqrt{x}+4}{\sqrt{x}-1}\le0\)

\(\Leftrightarrow\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}\le0\Rightarrow\sqrt{x}-1\le0\Leftrightarrow\sqrt{x}\le1\Leftrightarrow x\le1\)do \(\left(\sqrt{x}-2\right)^2\ge0\)

Kết hợp với đk, vậy \(x< 1\)

1, thay x=4 (TMĐKXĐ) vào P ta được:

P=\(\dfrac{1}{\sqrt{4}-1}\)=1

vậy khi x=4 thì P =1

2,với x≥0,x≠1:

Q=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)-\(\dfrac{2}{\sqrt{x}-1}-1\)=\(\dfrac{\sqrt{x}-2-\sqrt{x}+1}{\sqrt{x}-1}\)=\(\dfrac{-1}{\sqrt{x}-1}\)

vậy Q=\(\dfrac{-1}{\sqrt{x}-1}\)

3,\(\dfrac{1}{Q}+P\le4\)

⇒1/\(\dfrac{-1}{\sqrt{x}-1}\)+\(\dfrac{1}{\sqrt{x}-1}\)≤4⇔\(\dfrac{-\sqrt{x}-1}{1}+\dfrac{1}{\sqrt{x}-1}\le4\)⇔\(\dfrac{-x+1+1}{\sqrt{x}-1}-4\le0\)⇔\(\dfrac{-x+2-4\sqrt{x}+4}{\sqrt{x}-1}\le0\)⇔\(\dfrac{-x-4\sqrt{x}+6}{\sqrt{x}-1}\le0\)⇔\(\dfrac{x+4\sqrt{x}-6}{\sqrt{x}-1}\le0\)⇔\(\dfrac{x+4\sqrt{x}+4-10}{\sqrt{x}-1}\le0\)

\(\dfrac{ \left(\sqrt{x}+2\right)^2-10}{\sqrt{x}-1}\le0\)⇒\(\sqrt{x}-1\le0\) (vì (\(\sqrt{x}+2\))\(^2\)≥0 ∀ x hay (\(\sqrt{x}+2\))\(^2\)-10>0 ∀ x)

                                ⇔x≤1 (KTM)

vậy không có giá trị nào của x TM  để  \(\dfrac{1}{Q}+P\le4\)

☹