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 )

a, Với \(x\ge0;x\ne1\)

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

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

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

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

b, Ta có P = 3/4 hay \(\frac{2}{\sqrt{x}+1}=\frac{3}{4}\)ĐK : \(\sqrt{x}+1>0\)

\(\Rightarrow8=2\sqrt{x}+3\Leftrightarrow2\sqrt{x}=5\)

\(\Leftrightarrow\sqrt{x}=\frac{5}{2}\Leftrightarrow x=\frac{25}{4}\)

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

b, \(x= \dfrac{1}{9} ; x= 9\)

c, \(x= 0 ; x= 16\)

a) Với \(x>0;x\ne1\), ta có:

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

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

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

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

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

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

\(P=\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(\Rightarrow2P=\frac{2\sqrt{x}+2}{\sqrt{x}}\)

\(2P=2\sqrt{x}+5\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\left(ĐKXĐ:x\ne0\right)\left(1\right)\)

Mà theo đề bài : \(x>0\)nên phương trình luôn được xác định.

\(\left(1\right)\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=\frac{\sqrt{x}\left(2\sqrt{x}+5\right)}{\sqrt{x}}\)

\(\Rightarrow2\sqrt{x}+2=\sqrt{x}\left(2\sqrt{x}+5\right)\)

\(\Leftrightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)

\(\Leftrightarrow2\sqrt{x}+2-2x-5\sqrt{x}\)

\(\Leftrightarrow-2x-3\sqrt{x}+2=0\Leftrightarrow2x+3\sqrt{x}-2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{x}-1=0\\\sqrt{x}+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2\sqrt{x}=1\\\sqrt{x}=-2\left(vn\right)\end{cases}}\Leftrightarrow2\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(TMĐK:x>0;x\ne1\right)\)

Vậy \(2P=2\sqrt{x}+5\Leftrightarrow x=\frac{1}{4}\)

a/ \(P=12\)

b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:

\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )

a. Thay x = 3 vào biểu thức P ta được :

\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)

b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)

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

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

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

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

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

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

c, Ta có :

\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)

Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)

Trả lời:

a) Tính A khi x=9

Với x=9, A= \(\frac{\sqrt{9}}{\sqrt{9}-2}\)=3

b) Rút gọn:

T=A-B

T=\(\frac{\sqrt{x}}{\sqrt{x}-2}\)-\(\frac{2}{\sqrt{x}+2}\)-\(\frac{4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

T=\(\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

T=\(\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

c) Tìm x để T nguyên

T=\(\frac{\sqrt{x}-2}{\sqrt{x}+2}\)= 1-\(\frac{4}{\sqrt{x}+2}\)

T nguyên khi: 4 mod (\(\sqrt{x}\)+2)=0

=> \(\sqrt{x}+2\)={4,2,1}

=> \(\sqrt{x}\) ={2,0}

=> x={4,0}

Sao bài của mình làm khi post lên olm bị mất phần sau rồi ???

em làm luôn 

\(P=\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

b) thì em chưa làm đc :((

b, \(x=24-16\sqrt{2}=24-2.8.\sqrt{2}=24-8\sqrt{8}\)

\(=24-2.4\sqrt{8}=4^2-2.4\sqrt{8}+\left(\sqrt{8}\right)^2=\left(4-\sqrt{8}\right)^2\)

*, làm tiếp bước Q làm : \(\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(4-\sqrt{8}\right)^2}=\left|4-\sqrt{8}\right|=4-\sqrt{8}\)( vì \(4-\sqrt{8}>0\))

hay \(\frac{1}{4-\sqrt{8}-1}=\frac{1}{3-\sqrt{8}}=3+\sqrt{8}\)

Vậy với \(x=24-16\sqrt{2}\)thì \(P=3+\sqrt{8}\)