a) - Với \(x>0,x\ne1\), ta có:

\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)

\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)

\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)

\(A=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)

\(A=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)

\(A=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)

\(A=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)

\(A=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)

\(A=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)

Vậy với \(x>0,x\ne1\)thì \(A=\frac{1}{\sqrt{x}}\)

\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)

\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)

\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)

\(=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)

\(=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)

\(=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)

\(=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)

\(=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)

\(=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)

b) \(B=\left(x-\sqrt{x}+1\right)\cdot A=\frac{1}{\sqrt{x}}\left(x-\sqrt{x}+1\right)=\frac{x}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{x}}+\sqrt{x}-1\)

Xét hiệu B - 1 ta có : \(B-1=\frac{1}{\sqrt{x}}+\sqrt{x}-2=\frac{1}{\sqrt{x}}+\frac{x}{\sqrt{x}}-\frac{2\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)

Dễ thấy \(\hept{\begin{cases}\sqrt{x}>0\forall x>0\\\left(\sqrt{x}-1\right)^2\ge0\forall x\ge0\end{cases}}\Rightarrow\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\forall x>0\)

Đẳng thức xảy ra <=> x = 1 ( ktm ĐKXĐ )

Vậy đẳng thức không xảy ra , hay chỉ có B - 1 > 0 <=> B > 1 ( đpcm )

b) \(B=\left(x-\sqrt{x}+1\right).A\)

Với \(x>0.x\ne1\)thì \(B=\left(x-\sqrt{x}+1\right).\frac{1}{\sqrt{x}}=\frac{x-\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}-1+\frac{1}{\sqrt{x}}\)

\(B=\left(\sqrt{x}+\frac{1}{\sqrt{x}}-2\frac{\sqrt[4]{x}}{\sqrt[4]{x}}\right)+2\frac{\sqrt[4]{x}}{\sqrt[4]{x}}-1=\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2+2-1\)

\(B=\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2+1\)

Ta có:

\(\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2\ge0\forall x>0\)

Dấu bằng xảy ra \(\Leftrightarrow\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}=0\Leftrightarrow\sqrt[4]{x}=\frac{1}{\sqrt[4]{x}}\Leftrightarrow x=1\)(thỏa mãn điều kiện x>0)

Mà theo đề bài, \(x\ne1\)nên dấu bằng không xảy ra

Do đó : \(\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2>0\forall x\left(x>0;x\ne1\right)\)

\(\Rightarrow\left(\sqrt[4]{x}-\frac{1}{\sqrt[4]{x}}\right)^2+1>1\forall x\left(x>0;x\ne1\right)\)

\(\Rightarrow B>1\forall x\left(x>0;x\ne1\right)\)

Vậy với \(x>0;x\ne1\)thì \(B>1\)

a, \(A= \dfrac{4\sqrt{x}+2}{(\sqrt{x}+1)4\sqrt{x}}\)

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

a, A=\(\dfrac{1}{\sqrt{x}}\) với x>\(0;x\ne1\)                                                                                       b,B=\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\) >1 \(\Rightarrow\dfrac{x-\sqrt{x}+1-\sqrt{x}}{\sqrt{x}}>0\)  \(\Rightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\) \(\Rightarrow\left(\sqrt{x}-1\right)^2\) > 0\(\Rightarrow x>1\)  hay B>1                                                                          Vậy B>1 \(\forall x>0;x\ne1\)                                                                           

a/ ĐKXD : x > 0 , x≠1

A= \(\left(\dfrac{1}{x-1}+\dfrac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right).\left(x-1\right)}\right).\left(\dfrac{x+2\sqrt{x}+1}{4\sqrt{x}}-1\right)\)=\(\dfrac{\sqrt{x}-1+3\sqrt{x}+5}{\left(\sqrt{x}-1\right).\left(x-1\right)}.\dfrac{x-2\sqrt{x}+1}{4\sqrt{x}}=\dfrac{4.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)=\(\dfrac{1}{\sqrt{x}}\)

Vậy A=\(\dfrac{1}{\sqrt{x}}\) với x > 0 , x≠1

b/ Cho B=\(\left(x-\sqrt{x}+1\right).\dfrac{1}{\sqrt{x}}=\sqrt{x}-1+\dfrac{1}{\sqrt{x}}\)

$ADBDT\; cô\; si\; với\; 2\; số\; dương$

$\backslash (\backslash sqrt\{x\}+\backslash dfrac\{1\}\{\backslash sqrt\{x\}\}\backslash )$≥\(2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}\)=2

⇒\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\)-1≥1 

   hay B≥1 . Dấu "=" xảy ra ⇔ \(\sqrt{x}+\dfrac{1}{\sqrt{x}}\)⇒x=1 (TMDK)

 Vậy Bmin=1 khi x=1

a, A = \(\dfrac{4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(x-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)

   A = \(\dfrac{1}{\sqrt{x}}\)

b, Ta có :

\(\Rightarrow B=\left(x-\sqrt{x}-1\right).A\)

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

Ta có:

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

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

Mà \(\left(\sqrt{x}-1\right)^2>1\)

\(\Rightarrow x>1\left(B>1\right)\)

Vậy \(B>1\) với mọi x >0, x \(\ne1\)

Với x >0 ; x ≠1, ta có

A=\([\dfrac{1}{(\sqrt{x}-1)(\sqrt{x}+1)}+\dfrac{3\sqrt{x}+5}{(\sqrt{x}-1)^2(\sqrt{x}+1)}]\times\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{4\sqrt{x}}\)

A=\(\dfrac{\sqrt{x}-1+3\sqrt{x}+5}{(\sqrt{x}-1)^2(\sqrt{x}+1)}\times\dfrac{(\sqrt{x}-1)^2}{4\sqrt{x}}\)

A=\(\dfrac{4(\sqrt{x}+1)(\sqrt{x}-1)^2}{4(\sqrt{x}-1)^2(\sqrt{x}+1)\sqrt{x}}\)

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

Vậy với x >0 ;x≠1 thì A=\(\dfrac{1}{\sqrt{x}}\)

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

Áp dụng bất đẳng thức Cô si , ta có

\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\dfrac{\sqrt{x}}{\sqrt{x}}}\)

⇒\(\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\ge2-1=1\)

Dấu = xảy ra ⇔\(\dfrac{x+1}{\sqrt{x}}=2\Leftrightarrow2x+2=\sqrt{x}\)(Vô lý vì x>\(\sqrt{x}\) với∀ x>0,x≠1)

Vậy B>1 với mọi x>0 , x≠1

a) A= \(\dfrac{1}{\sqrt{x}}\)

b) \(B=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)= \(\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\)\(\ge2\sqrt{1}-1\)\(\ge1\)( Cosi)

​​​​​​​​​​

a,\(\left[\dfrac{1}{x-1}+\dfrac{3\sqrt{5}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right].\dfrac{\left(\sqrt{x}+1\right)^x-4\sqrt{x}}{4\sqrt{x}}\)

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

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

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

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

B>1

Áp dụng cô si  

\(=>\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\ge1\)

Mà x\(\ne\)1

=>B>1

B>1\(\forall x>0,x\ne1\)

a) A=\(\left(\dfrac{1}{x-1}+\dfrac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left(\dfrac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right)\)

=\(\left(\dfrac{1}{x-1}+\dfrac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right).\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{4\sqrt{x}}\)

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

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

b) Ta có

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

Áp dụng bất đẳng thức Cô si, ta có $B\ge 1$, đẳng thức xảy ra khi $x=1$.

Tuy nhiên, theo điều kiện  x\ne1x
​\(\ne\)1 nên $B>1$.

Vậy B>1\text{ }\forall x>0,x\ne1B>1 ∀x>0,x
​\(\ne\)1.

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

b) B > 1

a, A = \(\dfrac{1}{\sqrt{x}}\)với x > 0; x khác 1

b, Với x >0 và x khác 1 có :

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

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