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

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

b: Để A>-6 thì \(2\sqrt{x}< 6\)

=>0<x<9 

Kết hợp ĐKXĐ, ta được:

0<x<9 và x<>1

\(a,\)

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

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

\(=\dfrac{x-1}{2\sqrt{x}}.\dfrac{-4x}{x-1}\)

\(=-2\sqrt{x}\)

Vậy \(A=-2\sqrt{x}\)

\(b,\)Đề \(A\ge-6\) thì \(-2\sqrt{x}\ge-6\) \(\Leftrightarrow\sqrt{x}\le3\) \(\Leftrightarrow x\le3^2\Leftrightarrow x\le9\)

Vậy \(x\le9\) thì \(A\ge-6\)

a) ĐKXĐ: \(x>0;x\ne4\)

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

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

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

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

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

b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)

\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))

\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\) 

Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)

\(\text{#}\mathit{Toru}\)

\(a,A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\\ A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(b,\dfrac{P}{A}\left(x-1\right)=0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)

a) \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)

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

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

b) \(\dfrac{P}{A}\left(x-1\right)=0\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(x-1\right)=0\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow x=0\)( do \(\sqrt{x}+1\ge1>0\))(không thỏa đk)

Vậy \(S=\varnothing\)

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

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

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

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

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

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

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

mà \(\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}\left(\sqrt{x}-1\right)>0\)

mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}-1>0\)

\(\Leftrightarrow\sqrt{x}>1\)

hay x>1

Kết hợp ĐKXĐ, ta được: x>1

Vậy: Để P>0 thì x>1

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)ĐK:` \(\begin{cases}x \ge 0\\x-\sqrt{x} \ne 0\\x-1 \ne 0\\\end{cases}\)

`<=>` \(\begin{cases}x \ge 0\\x \ne 0\\x \ne 1\\\end{cases}\)

`<=>` \(\begin{cases}x>0\\x \ne 1\\\end{cases}\)

`b)A=(sqrtx/(sqrtx-1)-1/(x-sqrtx)):(1/(1+sqrtx)+2/(x-1))`

`=((x-1)/(x-sqrtx)):((sqrtx-1+2)/(x-1))`

`=(x-1)/(x-sqrtx):(sqrtx+1)/(x-1)`

`=(sqrtx+1)/sqrtx:1/(sqrtx-1)`

`=(x-1)/sqrtx`

`c)A>0`

Mà `sqrtx>0AAx>0`

`<=>x-1>0<=>x>1`

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

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

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

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

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

c, Ta có : \(A>0\)

\(\Leftrightarrow x-1>0\)

\(\Leftrightarrow x>1\)

Vậy ...

Lời giải:

ĐKXĐ: $x\geq 0; x\neq 1$

a)

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

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

b) 

\(A=7\Leftrightarrow x+\sqrt{x}+1=7(\sqrt{x}-1)\)

\(\Leftrightarrow x-6\sqrt{x}+8=0\Leftrightarrow (\sqrt{x}-2)(\sqrt{x}-4)=0\)

\(\Leftrightarrow \left[\begin{matrix} x=4\\ x=16\end{matrix}\right.\) (đều thỏa mãn)

c) 

\(x=2(2+\sqrt{3})=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\Rightarrow \sqrt{x}=\sqrt{3}+1\)

\(\Rightarrow A=\frac{4+2\sqrt{3}+\sqrt{3}+1+1}{\sqrt{3}}=\frac{6+3\sqrt{3}}{\sqrt{3}}=3+2\sqrt{3}\)

d)

\(A< 1\Leftrightarrow \frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1<0\Leftrightarrow \frac{x-2\sqrt{x}+2}{\sqrt{x}-1}<0\)

\(\Leftrightarrow \frac{(\sqrt{x}-1)^2+1}{\sqrt{x}-1}<0\Leftrightarrow \sqrt{x}-1< 0\Leftrightarrow 0\leq x< 1\)

đk : \(x\ge0,x\ne1\)

\(=>P=\left[\dfrac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]\)

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

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

b,\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\) thay vào P

\(=>P=\dfrac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}=\dfrac{2\sqrt{5}-3}{\sqrt{5}}\)

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

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

\(=>x-\sqrt{x}-\dfrac{1}{2}=>\Delta=1-4\left(-\dfrac{1}{2}\right)=3>0=>\left[{}\begin{matrix}x1=\dfrac{1+\sqrt{3}}{2}\\x2=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)

đối chiếu đk loại x2 còn x1 thỏa

1.\(x=4\)

\(B=\left(\dfrac{x+1}{2}-\sqrt{x}\right)=\left(\dfrac{4+1}{2}-\sqrt{4}\right)=\dfrac{5}{2}--2=\dfrac{5-4}{2}=\dfrac{1}{2}\)

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

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

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

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

3.\(M=\dfrac{\sqrt{x}}{6}\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{6}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=6\left(\sqrt{x}-1\right)\)

\(\Leftrightarrow x+\sqrt{x}=6\sqrt{x}-6\)

\(\Leftrightarrow x-5\sqrt{x}+6=0\)

Đặt \(\sqrt{x}=a;a\ge0\)

=> pt trở thành:

\(a^2-5a+6=0\)

\(\Delta=\left(-5\right)^2-4.6=25=24=1>0\)

=> pt có 2 nghiệm:

\(\left\{{}\begin{matrix}x_1=\dfrac{5+\sqrt{1}}{2}=3\left(tm\right)\\x_2=\dfrac{5-\sqrt{1}}{2}=2\left(tm\right)\end{matrix}\right.\)

Xét \(\sqrt{a}=3\)

\(\Leftrightarrow a=9\)

Xét \(\sqrt{a}=2\)

\(\Leftrightarrow a=4\)

Vậy \(x=9;4\)