a) Ta có:

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

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

\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)

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

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

b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)

.....Chưa nghĩ ra....

c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)

Vậy Min P = 0 khi x =9.

k - kb với tớ nhia mn!

a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

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

Sửa đề: \(A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac12\right)\)

ĐKXĐ: x>=0; x<>9

a: Ta có: \(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\)

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

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

\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=-\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

Ta có: \(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac12\)

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

Ta có: \(A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac12\right)\)

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

\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}=\frac{-6}{\sqrt{x}+3}\)

b: Để A nguyên thì -6⋮\(\sqrt{x}+3\)

=>\(\sqrt{x}+3\in\left\lbrace3;6\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;3\right\rbrace\)

=>x∈{0;9}

Kết hợp ĐKXĐ, ta được: x=0(nhận)

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

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

b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

hay \(x\in\left\{0;4;9\right\}\)

ĐKXĐ: x>0; x<>1

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

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

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

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

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

b: \(x=3+2\sqrt2=\left(\sqrt2+1\right)^2\)

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

\(=\frac{\sqrt{\left(\sqrt2+1\right)^2}+1}{\sqrt{\left(\sqrt2+1\right)^2}-1}=\frac{\sqrt2+1+1}{\sqrt2+1-1}\)

\(=\frac{2+\sqrt2}{\sqrt2}=\sqrt2+1\)

c: Để P nguyên thì \(\sqrt{x}+1\) ⋮\(\sqrt{x}-1\)

=>\(\sqrt{x}-1+2\) ⋮\(\sqrt{x}-1\)

=>2⋮\(\sqrt{x}-1\)

=>\(\sqrt{x}-1\in\left\lbrace1;-1;2;-2\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace2;0;3\right\rbrace\)

=>x∈{0;4;9}

Kết hợp ĐKXĐ, ta được: x∈{4;9}

a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\\x\ne9\end{cases}}\)

b) \(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x-3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}=\frac{-3\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}=-\frac{3}{2\left(\sqrt{x}-3\right)}\)c) Để P nguyên thì \(2\left(\sqrt{x}-3\right)\in\left\{-3;-1;1;3\right\}\)=> x thuộc rỗng.