ĐKXĐ: a>0
\(C=\left(\dfrac{1}{a+\sqrt{a}}+\dfrac{1}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}}{a+2\sqrt{a}+1}\)
\(=\left(\dfrac{1}{\sqrt{a}\left(\sqrt[]{a}+1\right)}+\dfrac{1}{\sqrt{a}+1}\right)\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
\(=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}}=\dfrac{\left(\sqrt{a}+1\right)^2}{a}\)
ĐKXĐ: x>0; x<>1
\(E=\left(\dfrac{x}{x-\sqrt{x}}+\dfrac{\sqrt{x}+1}{x-1}\right)\cdot\dfrac{\sqrt{x}\left(x-\sqrt{x}+1\right)}{x\sqrt{x}+1}\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{\sqrt{x}\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
ĐKXĐ: x>=0; x<>1
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2}{x-2\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
\(=\dfrac{2}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}-1}{2}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
ĐKXĐ: x>0; x<>4
\(D=\left(\dfrac{1}{3\sqrt{x}-6}+\dfrac{1}{x-2\sqrt{x}}\right):\left(\dfrac{1}{6}+\dfrac{1}{2\sqrt{x}}\right)\)
\(=\left(\dfrac{1}{3\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\dfrac{1}{2\sqrt{x}}+\dfrac{1}{2\cdot3}\right)\)
\(=\dfrac{\sqrt{x}+3}{3\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{3+\sqrt{x}}{6\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+3}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{6\sqrt{x}}{\sqrt{x}+3}=\dfrac{2}{\sqrt{x}-2}\)
ĐKXĐ: x>0; x<>4
\(F=\left(\dfrac{1}{x-2\sqrt{x}}-\dfrac{1}{x}\right)\cdot\dfrac{x-4}{6}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{1}{x}\right)\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{6}\)
\(=\dfrac{\sqrt{x}-\left(\sqrt{x}-2\right)}{x\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{6}\)
\(=\dfrac{2}{x}\cdot\dfrac{\left(\sqrt{x}+2\right)}{6}=\dfrac{\sqrt{x}+2}{3x}\)
ĐKXĐ: x>0; x<>1
\(Q=\left(\dfrac{x+1}{\sqrt{x}}-2\right)\cdot\left(\dfrac{x}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}=\dfrac{x-1}{\sqrt{x}}\)
mọi người giúp mik bài 4 câu a vs ạ 
