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

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

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

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

b/ Ta có: \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)

    Để \(P\in Z\) thì \(\left(\sqrt{x}+1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

    + Với \(\sqrt{x}+1=1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

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

    + Với \(\sqrt{x}+1=2\Rightarrow\sqrt{x}=1\Rightarrow x=1\)(loại)

    + Với \(\sqrt{x}+1=-2\Rightarrow\sqrt{x}=-3\left(vn\right)\)

                                         Vậy x = 0 thì P nguyên

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

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

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

Để P nguyên thì \(\sqrt{x}+1\in\left\{1;2\right\}\Leftrightarrow x\in\left\{0\right\}\) (Vì x khác 1 - điều kiện)

c) \(\sqrt{x}+1\ge1\Leftrightarrow\frac{2}{\sqrt{x}+1}\le\frac{1}{2}\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\ge\frac{1}{2}\)

\(\Rightarrow P\ge\frac{1}{2}\). Dấu đẳng thức xảy ra khi x = 0

Vậy Min P = 1/2 <=> x = 0

Sửa lại câu c) 

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

Ta có \(\sqrt{x}+1\ge1\Leftrightarrow\frac{2}{\sqrt{x}+1}\le2\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\le-1\)

Dấu "=" xảy ra khi x = 0

Vậy Min P = -1 khi và chỉ khi x = 0

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

Có \(\sqrt{x}+2\ge2\forall x\ge0\)

\(\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)\(\Rightarrow-\dfrac{3}{\sqrt{x}+2}\ge-\dfrac{3}{2}\)\(\Leftrightarrow1-\dfrac{3}{\sqrt{x}+2}\ge1-\dfrac{3}{2}=-\dfrac{1}{2}\)

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

Dấu "=" xảy ra khi x=0

Vậy \(min=-\dfrac{1}{2}\Leftrightarrow x=0\)

Q= \(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)+\(\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)= \(\frac{2\sqrt{x}-9-\left(x-9\right)+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)=\(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) Q <1 <=> \(\frac{\sqrt{x}-3+4}{\sqrt{x}-3}< 1< =>1+\frac{4}{\sqrt{x}-3}\)<1 <=> \(\frac{4}{\sqrt{x}-3}< 0\) <=> \(\sqrt{x}-3< 0< =>\sqrt{x}< 3\)<=> \(0\le\)x< 9

c) Q = 1 \(+\frac{4}{\sqrt{x}-3}\) là số nguyên khi 4 chia hết cho\(\sqrt{x}-3\) <=> \(\sqrt{x}-3=1;\sqrt{x}-3=-1;\sqrt{x}-3=2\);\(\sqrt{x}-3=-2;\sqrt{x}-3=4;\sqrt{x}-3=-4\)

<=> x= 16; x = 4; x = 25; x = 1 ; x = 49

Bài làm của bạn Mạnh có hai lỗi:

+) ĐKXĐ: \(\hept{\begin{cases}x-5\sqrt{x}+6\ne0;\sqrt{x}-2\ne0;3-\sqrt{x}\ne0\\x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne4;9\end{cases}}\)

+) Vì ko có điều kiện nên câu c chưa loại nghiệm. x = 4 loại nhé

\(\Leftrightarrow\)A=\(\left|x-2010\right|+\left|x-2011\right|\)=\(\left|x-2010\right|+\left|2011-x\right|\)\(\ge\)\(\left|x-2010+2011-x\right|\)=1

Dấu ''='' xảy ra khi và chỉ khi \(\hept{\begin{cases}x-2010\ge0\\2011-x\ge0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge2010\\x\le2011\end{cases}}\)\(\Leftrightarrow\)\(2010\le x\le2011\)

Vậy Min A =1 \(\Leftrightarrow2010\le x\le2011\)

chịu !!!

a) Với \(x\ge0;x\ne1\), ta có :

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

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

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

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

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

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

b) Ta có : P > 0

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

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)< 0\)

\(\Leftrightarrow\hept{\begin{cases}x\ne0\\\sqrt{x}-1< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\\sqrt{x}< 1\end{cases}\Leftrightarrow}}\hept{\begin{cases}x\ne0\\x< 1\end{cases}}\)

Kết hợp với đk đề bài , ta được 0 < x < 1

Vậy với 0 < x < 1 thì P > 0

c) Với \(x=7-4\sqrt{3}=3-2.2.\sqrt{3}+4=\left(\sqrt{3}-2\right)^2\)thì :

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

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

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

\(P=\sqrt{3}-3-3+2\sqrt{3}\)

\(P=3\sqrt{3}-5\)

Vậy với \(x=7-4\sqrt{3}\)thì \(P=3\sqrt{3}-5\)

d) Ta có \(P=-\sqrt{x}\left(\sqrt{x}-1\right)=\sqrt{x}-x=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\)

Nhận thấy : \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Rightarrow-\left(\sqrt{x}-\frac{1}{2}\right)^2\le0\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu " = " xảy ra khi và chỉ khi

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

Vậy với \(x=\frac{1}{4}\)thì max P là \(\frac{1}{4}\)

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

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

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

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

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

=> Với mọi \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)thì P = 2

Đề sai à --

kkk. thế mới hỏi chứ. đề đấy: đố giải được

a: ĐKXĐ: x>0; x<>1

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

c: A nguyên

=>x-1 thuộc {1;-1;2;-2}

=>x thuộc {2;3}

tự lm nhóe

Tự tìm ĐKXĐ nhé

\(P=\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)

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

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

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

\(=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

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

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

c, \(P=\frac{\sqrt{x}+4}{\sqrt{x}+2}=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)

Để \(P\in Z\Rightarrow1+\frac{2}{\sqrt{x}+2}\in Z\)

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

\(\Rightarrow\sqrt{x}=\left\{-1;0\right\}\)

\(\Rightarrow x=\left\{0\right\}\)

Kết hợp với ĐKXĐ =>...