a: \(x=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

\(\sqrt{x}-1\)

\(=\sqrt{\left(\sqrt5+1\right)^2}-1=\sqrt5+1-1=\sqrt5\)

\(2x+2-2x\cdot\sqrt{x}\)

\(=2\left(6+2\sqrt5\right)+2-2\cdot\left(6+2\sqrt5\right)\sqrt{\left(\sqrt5+1\right)^2}\)

\(=12+4\sqrt5+2-2\left(6+2\sqrt5\right)\left(\sqrt5+1\right)\)

\(=14+4\sqrt5-2\left(6\sqrt5+6+10+2\sqrt5\right)=14+4\sqrt5-2\left(8\sqrt5+16\right)\)

\(=14+4\sqrt5-16\sqrt5-32=-18-12\sqrt5\)

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

\(=\frac{\sqrt5}{-18-12\sqrt5}=\frac{-\sqrt5}{18+12\sqrt5}=\frac{-\sqrt5}{6\left(3+2\sqrt5\right)}=\frac{-\sqrt5\left(2\sqrt5-3\right)}{6\left(2\sqrt5+3\right)\left(2\sqrt5-3\right)}\)

\(=\frac{-10+3\sqrt5}{6\cdot\left(20-9\right)}=\frac{-10+3\sqrt5}{66}\)

a.

\(x=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\) \(\Rightarrow\sqrt{x}=\sqrt{5}+1\)

 \(\Rightarrow B=\dfrac{\sqrt{5}+1-1}{2+\sqrt{5}+1}=\dfrac{\sqrt{5}}{\sqrt{5}+3}=\dfrac{3\sqrt{5}-5}{4}\)

b.

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

B nguyên \(\Rightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Rightarrow\sqrt{x}+2=Ư\left(3\right)\)

Mà \(\sqrt{x}+2\ge2\Rightarrow\sqrt{x}+2=3\)

\(\Leftrightarrow\sqrt{x}=1\Rightarrow x=1\)

Mình cảm ơn.

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

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

\(Q=x+1\)

Không thể tìm được GTLN hay GTNN của Q.

b)

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

Để \(\frac{3Q}{\sqrt{x}}\) nguyên thì \(\frac{3}{\sqrt{x}}\)nguyên hay \(\sqrt{x}\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Vì \(\sqrt{x}\)dương nên \(\sqrt{x}\in\left\{1;3\right\}\)

Vậy x=1, x=9 là các giá trị cần tìm

P=A*B

\(=\dfrac{x-7}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{x-7}{\sqrt{x}+2}\)

P nguyên

=>x-4-3 chia hết cho căn x+2

=>căn x+2 thuộc Ư(-3)

=>căn x+2=3

=>x=1

a: Thay \(x=\dfrac{1}{4}\) vào A, ta được:

\(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-2\right)=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)

b: Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)

\(=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{\sqrt{x}+4}{\sqrt{x}-2}\)

c: Để B là số tự nhiên thì \(\sqrt{x}+4⋮\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{1;2;3;6\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{3;4;5;8\right\}\)

hay \(x\in\left\{16;25;64\right\}\)

a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)

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

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

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

a. B = \(\dfrac{\sqrt{36}}{\sqrt{36}-3}=\dfrac{6}{6-3}=2\)

a: Thay x=36 vào B, ta được:

\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)