Lời giải:

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

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

a, ĐK: \(x>0\)

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

b, ĐK: \(x\ge0;x\ne1\)

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

ĐKXĐ: \(x\ge0;x\ne3\)

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

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

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

g: \(\dfrac{\sqrt{x}+3}{x\sqrt{x}+27}=\dfrac{1}{x-3\sqrt{x}+9}\)

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

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

k: \(\dfrac{x+7\sqrt{x}+12}{x-9}=\dfrac{\sqrt{x}+4}{\sqrt{x}-3}\)

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

ĐKXĐ: \(x\ge0;x\ne4\)

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

\(=\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}-\dfrac{12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

\(D=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3\sqrt{x}+1}{x-1}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(x\ge0;x\ne1\right)\\ D=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\\ D=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

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

\(A=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0;x\ne25\right)\\ A=\dfrac{15-\sqrt{x}+2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\\ A=\dfrac{5+\sqrt{x}}{\sqrt{x}+5}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+1}\)

\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}+7}{x-4}\right)\)

\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)

\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\)

\(=\dfrac{-x+8\sqrt{x}-15+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{-x+8\sqrt{x}-15+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(ĐK:x\ge0;x\ne4\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}-5\right)+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{8\sqrt{x}-15-x+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

Ta có: \(B=\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-5}{x-4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(=\dfrac{2\sqrt{x}-4-\sqrt{x}+5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

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

\(B=\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-5}{x-4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(x\ge0;x\ne4\right)\\ B=\dfrac{2\sqrt{x}-4-\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\\ B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+2}\)