\(B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\cdot\dfrac{\sqrt{x}-2+3}{3}\)

\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}+1}{3}=\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}\)

Ta có: M=A+B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

\(B=\dfrac{9\sqrt{x}+15-3\sqrt{x}+3-4\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

\(B=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-\dfrac{36}{x-9}\)

\(=\dfrac{\left(\sqrt{x}+3\right)^2+\left(\sqrt{x}-3\right)^2-36}{x-9}\)

\(=\dfrac{x+6\sqrt{x}+9+x-6\sqrt{x}+9-36}{x-9}\)

\(=\dfrac{2x-18}{x-9}=\dfrac{2\left(x-9\right)}{x-9}=2\)

ĐKXĐ : \(x\ge0;x\ne9\)

Ta có : \(B=\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\dfrac{x+6\sqrt{x}+9+x-6\sqrt{x}+9-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=2\)

b, tìm x thuộc Z để B thuộc Z

c, Tìm x thuộc R để B có giá trị nguyên

\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

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

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

\(\Leftrightarrow B=\left(1-\frac{\sqrt{x}}{\sqrt{x}+3}\right):\frac{x-9-x+9+x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

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

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

\(\Leftrightarrow B=\frac{3}{\sqrt{x}+3}:\frac{\sqrt{x}+2}{\sqrt{x}+3}\)

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

\(\Leftrightarrow B=\frac{3}{\sqrt{x}+2}\)

b) ??

b) Để \(B\inℤ\)

\(\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1\right\}\)(Loại các giá trị âm)

\(\Leftrightarrow x=1\)

a: Khi x=16 thì B=1/(4-3)=1

b: P=A-B

\(=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{x-9}=\dfrac{x+\sqrt{x}-6}{x-9}=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\)

ĐK: \(x\ge0;x\ne9\)

a) Khi \(x=16\) TMĐKXĐ thì \(B=\dfrac{1}{\sqrt{16}-3}=1\)

b) \(P=A-B\)

\(P=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

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

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

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

c) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

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

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

\(\Leftrightarrow x+2\sqrt{x}+2\sqrt{x}+4=x+3\sqrt{x}+\sqrt{x}+3\)

\(\Leftrightarrow4=3\) (Sai)

Vậy \(x\in\varnothing\)

\(a,x=16\Rightarrow B=\dfrac{1}{\sqrt{16}-3}=\dfrac{1}{4-3}=1\)

\(b,\) Rút gọn : \(A=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}\left(dkxd:x\ne9,x\ge0\right)\)

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

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

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

\(=\dfrac{x+2\sqrt{x}-3}{x-9}\)

  Rút gọn \(P=A-B=\dfrac{x+2\sqrt{x}-3}{x-9}-\dfrac{1}{\sqrt{x}-3}\left(dkxd:x\ge0,x\ne9\right)\)

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

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

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

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

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

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

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

\(c,P=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\left(dkxd:x\ne9,x\ne4,x\ge0\right)\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-3}=0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=0\)

\(\Leftrightarrow\dfrac{x-3\sqrt{x}+\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=0\)

\(\Leftrightarrow-2\sqrt{x}+1=0\) ( Mất mẫu là bạn lấy mẫu nhân ngược vào 0 bên vế phải nha. )

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

\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

Vậy khi \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) thì \(x=\dfrac{1}{4}\)