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

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

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

\(P_{Min}=-\frac{1}{4}\) khi \(\sqrt{x}=\frac{3}{2}\Leftrightarrow x=\frac{9}{4}\)

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

\(Q\ge\frac{1}{\frac{\left(\sqrt{x}+1+2-\sqrt{x}\right)^2}{4}}=\frac{4}{3^2}=\frac{4}{9}\)

\(Q_{min}=\frac{4}{9}\) khi \(\sqrt{x}+1=2-\sqrt{x}\Leftrightarrow x=\frac{1}{4}\)

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

Chắc là bạn ghi nhầm đề, với \(x< 1\) biểu thức này ko có min

Nó chỉ có min khi \(x>1\)

Khi đó: \(R=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\ge2\sqrt{\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}}+3=3+2\sqrt{2}\)

\(R_{min}=3+2\sqrt{2}\) khi \(\sqrt{x}-1=\sqrt{2}\Leftrightarrow x=3+2\sqrt{2}\)

a) ĐKXĐ \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

b) ĐKXĐ: \(\left\{{}\begin{matrix}a,b\ge0\\a\ne b\end{matrix}\right.\)

\(\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}+2\sqrt{ab}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}+2\sqrt{ab}=\sqrt{a}-\sqrt{b}+2\sqrt{ab}\)

bài 1:

a) \(m>1\)

=>\(\sqrt{m}>\sqrt{1}\)

=>\(\sqrt{m}>1\)

b) \(m< 1\)

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

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

\(P=\left(\frac{1}{\sqrt{a}+2}+\frac{1}{\sqrt{a}-2}\right).\frac{\sqrt{a}-2}{\sqrt{a}}\)(ĐK: \(a>0\) và \(a\ne4\))

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

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

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

\(=\frac{2}{\sqrt{a}+2}\)

a)Để P>1/3 thì

\(\frac{2}{\sqrt{a}+2}>\frac{1}{3}\)

\(\Leftrightarrow\sqrt{a}+2< 6\)

\(\Leftrightarrow\sqrt{a}< 4\)

\(\Leftrightarrow a< 16\)

Kết hợp với đkxđ ta được \(0< a< 16\) và \(a\ne4\) thì P>1/3

b) Ta có:

\(Q=\frac{9}{2}P=\frac{9}{2}.\frac{2}{\sqrt{a}+2}=\frac{9}{\sqrt{a}+2}\)

Để Q nguyên thì \(9⋮\left(\sqrt{a}+2\right)\)

\(\Rightarrow\sqrt{a}+2\in\left\{-9;-3;-1;1;3;9\right\}\)

\(\Rightarrow\sqrt{a}\in\left\{-11;-5;-3;-1;1;7\right\}\)

\(\Rightarrow a\in\left\{1;49\right\}\)

Bài 1.

\(B=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\div\frac{x}{x-\sqrt{x}}\)với \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

a) \(B=\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)

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

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

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

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

\(B=\frac{4\sqrt{x}\cdot\sqrt{x}}{\left(\sqrt{x}+1\right)x}=\frac{4x}{\left(\sqrt{x}+1\right)x}=\frac{4}{\sqrt{x}+1}\)

b) Để B > 1

=> \(\frac{4}{\sqrt{x}+1}>0\)( với \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\))

Vì 4 > 0

=> \(\sqrt{x}+1>0\)

<=> \(\sqrt{x}>-1\)( luôn luôn đúng \(\forall\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)) ( theo ĐKXĐ )

Vậy \(\forall\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)thì B > 1

Chưa chắc lắm ... Còn câu 2 thì tí nữa mình làm cho 

Bài 2.

\(A=2\sqrt{5}-1\)

\(B=\frac{2}{x-1}\cdot\sqrt{\frac{x^2-2x+1}{4x^2}}\)( x > 0 )

a) \(B=\frac{2}{x-1}\cdot\frac{\sqrt{x^2-2x+1}}{\sqrt{4x^2}}\)

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

\(B=\frac{2}{x-1}\cdot\frac{\left|x-1\right|}{\left|2x\right|}\)

\(B=\frac{2}{x-1}\cdot\frac{x-1}{2x}=\frac{1}{x}\)( vì x > 0 )

b) Để A + B = 0

=> \(\left(2\sqrt{5}-1\right)+\frac{1}{x}=0\)( ĐKXĐ : \(x\ne0\))

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

<=> \(\frac{1}{x}=1-2\sqrt{5}\)

<=> \(x\times\left(1-2\sqrt{5}\right)=1\)

<=> \(x=\frac{1}{1-2\sqrt{5}}\)( tmđk )

Vậy \(x=\frac{1}{1-2\sqrt{5}}\)

a) Thay x=25 vào biểu thức \(A=\frac{7}{\sqrt{x}+8}\), ta được:

\(A=\frac{7}{\sqrt{25}+8}=\frac{7}{5+8}=\frac{7}{13}\)

Vậy: khi x=25 thì \(A=\frac{7}{13}\)

b) Ta có: \(B=\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{2\sqrt{x}-24}{x-9}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x+8\sqrt{x}-3\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+8\right)-3\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)

c) Ta có: \(P=A\cdot B\)

\(=\frac{7}{\sqrt{x}+8}\cdot\frac{\sqrt{x}+8}{\sqrt{x}+3}=\frac{7}{\sqrt{x}+3}\)

ĐKXĐ: \(x\ge0\)

Để P có giá trị nguyên thì \(7⋮\sqrt{x}+3\)

\(\Leftrightarrow\sqrt{x}+3\inƯ\left(7\right)\)

\(\Leftrightarrow\sqrt{x}+3\in\left\{1;-7;-1;7\right\}\)

\(\Leftrightarrow\sqrt{x}+3=7\)(vì \(\sqrt{x}+3\ge3\forall x\ge0\))

\(\Leftrightarrow\sqrt{x}=4\)

hay x=16(nhận)

Vậy: Khi x=16 thì P nguyên

d) Ta có: \(\sqrt{x}+3\ge3\forall x\ge0\)

\(\Leftrightarrow\frac{7}{\sqrt{x}+3}\le\frac{7}{3}\forall x\ge0\)

Dấu '=' xảy ra khi x=0

Vậy: Giá trị lớn nhất của biểu thức \(P=A\cdot B\) là \(\frac{7}{3}\) khi x=0

e) Để \(P=\frac{1}{2}\) thì \(\frac{7}{\sqrt{x}+3}=\frac{1}{2}\)

\(\Leftrightarrow\sqrt{x}+3=7\cdot2=14\)

\(\Leftrightarrow\sqrt{x}=14-3=11\)

hay x=121(nhận)

Vậy: để \(P=\frac{1}{2}\) thì x=121