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

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

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

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

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

b: Khi x=9/4 thì \(P=\dfrac{3}{2}:\left(\dfrac{3}{2}-1\right)=\dfrac{3}{2}:\dfrac{1}{2}=3\)

c: P<0

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

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

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

=>0<=x<1

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:

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

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

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

\(a,A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\left(dk:x\ge0,x\ne4\right)\\ =\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right)\\ =\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x-4+10-x}\)

\(=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\sqrt{x}-2}.\dfrac{1}{6}\\ =\dfrac{-6}{\left(\sqrt{x}-2\right).6}\\ =-\dfrac{1}{\sqrt{x}-2}\)
\(b,A>0\Leftrightarrow-\dfrac{1}{\sqrt{x}-2}>0\Leftrightarrow\sqrt{x}-2< 0\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với \(dk:x\ge0,x\ne4\), ta kết luận \(0\le x< 4\)

Mình cần gấp nhớ đừng làm tắt nhé 

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

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

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

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

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

b/

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

Để B nguyên

\(x-1=\left\{-1;-2;1;2\right\}\Rightarrow x=\left[0;-1;2;3\right]\)

tìm cả đk giúp mik vs

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

b.

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

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

c.

Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)

Ta có:

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

Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)

a: \(A=3+\left(-2\right)\cdot\sqrt{3}+3\cdot\sqrt{3}-2-\sqrt{3}\)

\(=3-2=1\)

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

b: B<A

=>B-1<0

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

=>-1/căn x<0

=>căn x>0

=>x>0 và x<>1

*Rút gọn

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

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

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

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

Ta có: \(C=x-\sqrt{x}+1\)

\(=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{x}=\dfrac{1}{2}\)

hay \(x=\dfrac{1}{4}\)

\(C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\left(x>0;x\ne1\right)\)

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

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

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

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

Dấu "=" xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}\)

Vậy  \(C_{min}=\dfrac{3}{4}\)

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

Áp dụng AM-GM có: \(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\)

Dấu "=" xảy ra khi x=1 (ktm đk)

Suy ra dấu bằng ko xảy ra \(\Rightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}-1>2-1=1\)

\(\Rightarrow\dfrac{2}{\sqrt{x}+\dfrac{1}{\sqrt{x}}-1}< 2\) 

\(\Rightarrow N< 2\) mà \(N>0\),\(N\) nguyên

\(\Rightarrow N=1\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{5}}{2}\\\sqrt{x}=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\\x=\dfrac{7-3\sqrt{5}}{2}\end{matrix}\right.\) (tm)

Vậy...

a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

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

\(=-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

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

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

b) \(A>B\Rightarrow-\sqrt{7}>\dfrac{8}{\sqrt{x}-3}\Rightarrow\dfrac{8}{\sqrt{x}-3}+\sqrt{7}< 0\)

\(\Rightarrow\dfrac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}< 0\)

Ta có: \(\left\{{}\begin{matrix}8=\sqrt{64}\\3\sqrt{7}=\sqrt{63}\end{matrix}\right.\Rightarrow8-3\sqrt{7}>0\Rightarrow8-3\sqrt{7}+\sqrt{7x}>0\)

\(\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0< x< 9\)