1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)

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

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

3: 2P=2*căn x+5

=>\(\dfrac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\)

=>\(2x+5\sqrt{x}-2\sqrt{x}-2=0\)

=>\(2x+3\sqrt{x}-4=0\)

=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)

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

=>x=1/4

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

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

a, x > 0 ; x khác 1 

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

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

b, Ta có : \(P=\dfrac{x-2}{\sqrt{x}}=1\Rightarrow x-2=\sqrt{x}\)

\(\Leftrightarrow x-\sqrt{x}-2=0\Leftrightarrow\left(\sqrt{x}+1>0\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow x=4\)(tm) 

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

b: Để P=1 thì \(x-\sqrt{x}-2=0\)

hay x=4

Câu b bạn sửa lại đề

\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)

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

Lời giải:
1. \(P=\left[\frac{1}{\sqrt{x}(\sqrt{x}-1)}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}\right]:\frac{x}{(\sqrt{x}-1)^2}\)

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

2.

\(P>\frac{1}{2}\Leftrightarrow \frac{x-1}{x\sqrt{x}}> \frac{1}{2}\)

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

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

\(\Leftrightarrow x\sqrt{x}+2< 2x\) 

Điều này vô lý do theo BĐT Cô-si thì:\(x\sqrt{x}+2=\frac{x\sqrt{x}}{2}+\frac{x\sqrt{x}}{2}+2\geq 3\sqrt[3]{\frac{x^3}{2}}>\frac{3x}{\sqrt[3]{2}}> 2x\)

Vậy không tồn tại $x$ thỏa mãn.

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

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

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

a.

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

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

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

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

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

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

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

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

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

A = \(\dfrac{-3x-4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

A = \(\dfrac{-\left(3x+4\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

A = \(\dfrac{-\sqrt{x}\left(3\sqrt{x}+4\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

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

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

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

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

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

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

b: Để P>0 thì \(-\dfrac{\sqrt{x}}{\sqrt{x}-1}>0\)

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

=>\(\sqrt{x}< 1\)

=>\(0< =x< 1\)

c: Thay \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) vào P, ta được:

\(P=\dfrac{-\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-1}\)

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

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

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

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

\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)

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

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

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

Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)

b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)

hay \(P>0\forall x>0,x\ne1\)(đpcm)