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.

a ) \(ĐKXĐ:x\ge0;x\ne1\)

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

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

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

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

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

B ) Ta có :

 \(Q=P-\sqrt{x}\)

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

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

Đế Q nhận giá trị nguyên thì \(1+\frac{3}{\sqrt{x}-1}\in Z\)

\(\Leftrightarrow\frac{3}{\sqrt{x}-1}\in Z\left(vì1\in Z\right)\)

\(\Leftrightarrow\sqrt{x}-1\inƯ\left(3\right)\)

Ta có bảng sau :

Vậy để biểu thức \(Q=P-\sqrt{x}\) nhận giá trị nguyên thì \(x\in\left\{16;4;0\right\}\)


 

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

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

Phương trình tương đương : 

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

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

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

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

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

b) 

 Để \(A\in Z\)

\(\Rightarrow1+\sqrt{x}\inƯ\left(-2\right)\)

\(\Rightarrow\sqrt{x}\in\left\{0;-2;1;-3\right\}\)mà \(\sqrt{x}\ge0\)

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

\(\Rightarrow x\in\left\{0;1\right\}\)

a, \(M=\frac{\sqrt{x}}{\sqrt{x}+6}+\frac{1}{\sqrt{x}-6}+\frac{17\sqrt{x}+30}{\left(\sqrt{x}+6\right)\left(\sqrt{x}-6\right)}\)

\(=\frac{x-6\sqrt{x}+\sqrt{x}+6+17\sqrt{x}+30}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\frac{12\sqrt{x}+x+36}{\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-6}\)

b, Ta có : \(L=N.M\Rightarrow L=\frac{\sqrt{x}+6}{\sqrt{x}-6}.\frac{24}{\sqrt{x}+6}=\frac{24}{\sqrt{x}+6}\)

Vì \(\sqrt{x}+6\ge6\)

\(\Rightarrow\frac{24}{\sqrt{x}+6}\le\frac{24}{6}=4\)

Dấu ''='' xảy ra khi \(\sqrt{x}+6=6\Leftrightarrow x=0\)

Vậy GTLN L là 4 khi x = 0