Bài 5: 

a: Thay \(x=4+2\sqrt{3}\) vào E, ta được:

\(E=\dfrac{\sqrt{3}+1-1}{\sqrt{3}+1-3}=\dfrac{\sqrt{3}}{\sqrt{3}-2}=-3-2\sqrt{3}\)

b: Để E<1 thì E-1<0

\(\Leftrightarrow\dfrac{\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\Leftrightarrow\sqrt{x}-3< 0\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)

c: Để E nguyên thì \(4⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7\right\}\)

hay \(x\in\left\{16;25;49\right\}\)

Câu 2:
a) Ta có \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)

Thay \(x=\sqrt{3}-1\) vào \(B\), ta được

\(B=\dfrac{\sqrt{3}-1-2}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-3}{\sqrt{3}}=1-\sqrt{3}\)

b) Để \(B\) âm thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\) mà \(\sqrt{x}+1\ge1>0\forall x\) \(\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)

c) Ta có \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)

Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{3}{\sqrt{x}+1}\le3\Rightarrow B=1-\dfrac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow x=0\)

Vậy \(B_{min}=-2\) khi \(x=0\)

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

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

1) Sửa đề: x=0,09

Thay x=0,09 vào A, ta được:

\(A=\dfrac{\sqrt{0.09}}{\sqrt{0.09}-1}=\dfrac{0.3}{0.3-1}=\dfrac{0.3}{-0.7}=\dfrac{-3}{7}\)

\(a,x=16\Rightarrow A=\dfrac{\sqrt{16}+2}{\sqrt{16}-3}=\dfrac{4+2}{4-3}=6\)

\(b,B=\dfrac{\sqrt{x}+5}{\sqrt{x}+1}+\dfrac{\sqrt{x}-7}{1-x}\left(dk:x\ge0,x\ne1,x\ne9\right)\\ =\dfrac{\sqrt{x}+5}{\sqrt{x}+1}-\dfrac{\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-7\right)}{x-1}\\ =\dfrac{x+4\sqrt{x}-5-\sqrt{x}+7}{x-1}\\ =\dfrac{x+3\sqrt{x}+2}{x-1}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(dpcm\right)\)

\(c,\dfrac{4A}{A}\le\dfrac{x}{\sqrt{x}-3}\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}:\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\le\dfrac{x}{\sqrt{x}-3}\)

\(\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\le\dfrac{x}{\sqrt{x}-3}\)

\(\Leftrightarrow4-\dfrac{x}{\sqrt{x}-3}\le0\)

\(\Leftrightarrow\dfrac{4\sqrt{x}-12-x}{\sqrt{x}-3}\le0\)

\(\Leftrightarrow\) Pt vô nghiệm

Vậy không có giá trị x thỏa yêu cầu đề bài.

a. B = \(\dfrac{\sqrt{36}}{\sqrt{36}-3}=\dfrac{6}{6-3}=2\)

a: Thay x=36 vào B, ta được:

\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)

a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{-3}{\sqrt{x}+3}\)

b) Ta có: \(x=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

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

=2

Thay x=2 vào A, ta được:

\(A=\dfrac{-3}{3+\sqrt{2}}=\dfrac{-9+3\sqrt{2}}{7}\)

a, \(\Rightarrow M=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

 \(\Rightarrow M=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

b, \(x=3+2\sqrt{2}\Rightarrow M=\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}=\dfrac{\sqrt{2+2\sqrt{2}.1+1}-2}{\sqrt{2+2\sqrt{2}.1+1}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2-2\sqrt{2}+1}{2-1}=3-2\sqrt{2}\)

c, \(M>0\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\Rightarrow x>4\)

ĐK: \(x-9\ne0\Rightarrow x\ne9\)

\(\sqrt{x}\ge0\Rightarrow x\ge0\)

\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)

\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)

ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)

\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)

2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)

\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)

\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)