ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

a) Vì \(-1< 0\) nên không tính được A

a) Vì \(x\ne1\) nên không tính được A

a: \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

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

\(=\sqrt{x}-1\)

a) \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

                        Đk: \(x>0\) và \(x\ne1\)

\(\Rightarrow A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)

        \(=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

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

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

b) Thay \(x=3+2\sqrt{2}\) vào A ta được:

  \(A=\sqrt{3+2\sqrt{2}}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1\)

      \(=\sqrt{2}+1-1=\sqrt{2}\)

(Vì \(\sqrt{2}+1>0\Rightarrow\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\))

a: \(B=\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{4-x}\)

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

\(=\dfrac{\sqrt{x}+2+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

Khi x=16 thì \(B=\dfrac{2\cdot4+2}{\left(4-2\right)\left(4+2\right)}=\dfrac{10}{2\cdot6}=\dfrac{10}{12}=\dfrac{5}{6}\)

b: P=B/A

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

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

c: P<1

=>P-1<0

=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)

=>\(\dfrac{3}{\sqrt{x}-2}< 0\)

=>\(\sqrt{x}-2< 0\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

mà x nguyên

nên \(x\in\left\{0;1;2;3\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;1;2;3\right\}\)

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

\(A=\sqrt{x}\cdot\sqrt{x}+\sqrt{x}-\left(x-2\sqrt{x}+1\right)-2\)

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

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

\(A=3\sqrt{x}-3\)

Thay \(x=9\) vào A ta có:

\(A=3\cdot\sqrt{9}-3=3\cdot3-3=9-3=6\)

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}

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

Mà \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow B=\left(\sqrt{3}-1-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

\(b,P=AB=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\\ P=\dfrac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}=\sqrt{x}-1\\ c,Q=\sqrt{x}+\dfrac{1}{P}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}\\ Q=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+1\ge2\sqrt{1}+1=3\\ Q_{min}=3\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=1\\1-\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\sqrt{x}=2\left(x>1\Leftrightarrow\right)x=4\left(tm\right)\)

a: \(B=\left(\sqrt{x}-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

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

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 1; x\neq 25$

a) 

\(A=\frac{4\sqrt{x}}{\sqrt{x}-5}:\left[\frac{(\sqrt{x}-2)(\sqrt{x}+2)+\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+2}+\frac{5-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}\right]\)

\(=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{x-4+\sqrt{x}-1+5-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+2)}\)

\(=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+2)}=\frac{4\sqrt{x}}{\sqrt{x}-5}:\frac{\sqrt{x}}{\sqrt{x}+2}=\frac{4\sqrt{x}}{\sqrt{x}-5}.\frac{\sqrt{x}+2}{\sqrt{x}}=\frac{4(\sqrt{x}+2)}{\sqrt{x}-5}\)

b) Tại $x=81$ thì $\sqrt{x}=9$.

Khi đó: $A=\frac{4(9+2)}{9-5}=11$

c) $A< 4\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}-5}< 1$

$\Leftrightarrow \frac{7}{\sqrt{x}-5}< 0\Leftrightarrow \sqrt{x}-5< 0$

$\Leftrightarrow 0\leq x< 25$. Kết hợp với ĐKXĐ suy ra: $0\leq x< 25; x\neq 1$

\(ĐK:x\ne-1\\ \left|x\right|=2\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

Với \(x=2\Leftrightarrow A=\dfrac{3}{2+1}=1\)

Với \(x=-2\Leftrightarrow A=\dfrac{3}{-2+1}=-3\)