a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)

\(=4-2\sqrt{3}+2\sqrt{3}\)

=4

Thay x=4 vào B, ta được:

\(B=\dfrac{2-4}{2}=-1\)

Bài `1`

\(\sqrt{4-2\sqrt{3}}-\dfrac{2}{\sqrt{3}+1}+\dfrac{\sqrt{3}-3}{\sqrt{3}-1}\\ =\sqrt{3-2\sqrt{3}+1}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}+1-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}+1-\sqrt{3}\\ =-\sqrt{3}\)

2:

a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)

b: B=5

=>\(5\left(\sqrt{x}+3\right)=\sqrt{x}+8\)

=>\(5\sqrt{x}+15=\sqrt{x}+8\)

=>\(4\sqrt{x}=-7\)(loại)

Vậy: \(x\in\varnothing\)

đề hơi sai, sửa này mới đúng nhaa

a) \(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

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

b, x = \(4-\sqrt{12}\) = \(\left(\sqrt{3}-1\right)^2\) => \(\sqrt{x}=\sqrt{3}-1\) (1)

Thay (1) vào B, ta được : \(B=\dfrac{\sqrt{3}-1-1}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-2}{\sqrt{3}}\)

c, Để \(\sqrt{x}+1\ge2x-2\sqrt{x}-3\)

<=> \(2x-3\sqrt{x}-4\le0\)

xem lại đề hoặc nếu đề chuẩn rồi í thì c pt thành nhân tử rồi lấy trong khoảng (có lấy dấu bằng) =(( chứ đà này chuẩn bị rối

Với `x >= 0,x \ne 1` có:

`B=[\sqrt{x}+3]/[\sqrt{x}+1]-3/[1-\sqrt{x}]+4/[x-1]`

`B=[(\sqrt{x}+3)(\sqrt{x}-1)+3(\sqrt{x}+1)+4]/[(\sqrt{x}-1)(\sqrt{x}+1)]`

`B=[x-\sqrt{x}+3\sqrt{x}-3+3\sqrt{x}+3+4]/[(\sqrt{x}-1)(\sqrt{x}+1)]`

`B=[x+5\sqrt{x}+4]/[(\sqrt{x}-1)(\sqrt{x}+1)]`

`B=[(\sqrt{x}+4)(\sqrt{x}+1)]/[(\sqrt{x}-1)(\sqrt{x}+1)]`

`B=[\sqrt{x}+4]/[\sqrt{x}-1]`

\(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-\dfrac{3}{1-\sqrt{x}}+\dfrac{4}{x-1}\)

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

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

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

\(B=\dfrac{x+5\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(B=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(B=\dfrac{\sqrt{x}+4}{\sqrt{x}-1}\)

\(a,M=\left(\dfrac{\sqrt{x}+2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{x+3}{x-1}\right)\\ =\left(\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}+1-\sqrt{x}\left(\sqrt{x}-1\right)+x+3}{x-1}\right)\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+1-x+\sqrt{x}+x+3}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{2\sqrt{x}+4}\)

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

`b,` Để `M>1` Thì :

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

\(\Leftrightarrow-\sqrt{x}+1>0\) `(` Vì \(2\sqrt{x}>0\)  do \(x>0\) `)`

\(\Leftrightarrow-\sqrt{x}>-1\\ \Rightarrow x< 1\)

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{10}{8}=\dfrac{5}{4}\)

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

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

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

3: A/B>3/2

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{3}{2}>0\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

=>\(\dfrac{2\sqrt{x}+2-3\sqrt{x}}{\sqrt{x}\cdot2}>0\)

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

=>-căn x>-2

=>căn x<2

=>0<x<4

1) Thay x=64 vào A ta có:

\(A=\dfrac{2+\sqrt{64}}{\sqrt{64}}=\dfrac{2+8}{8}=\dfrac{5}{4}\)

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

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

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

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

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

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

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

3) Ta có:

\(\dfrac{A}{B}>\dfrac{3}{2}\) khi

\(\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}>\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{3}{2}>0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)

\(\Leftrightarrow\dfrac{2-\sqrt{x}}{2\sqrt{x}}>0\)

Mà: \(2\sqrt{x}\ge0\forall x\)

\(\Leftrightarrow2-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 2\)

\(\Leftrightarrow x< 4\)

Kết hợp với đk:

\(0< x< 4\)

Bài 2: 

Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)