a, ĐKXĐ: \(xe1;x\ge0\)
b, \(C=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{1-\sqrt{x}}\)
\(\Leftrightarrow C=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(\Leftrightarrow C=\dfrac{3\left(x+\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow C=\dfrac{3x+3\sqrt{x}-3-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow C=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
c, \(C=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1-2}{\sqrt{x}-1}=1-\dfrac{2}{\sqrt{x}-1}\)
Để C nguyên thì \(\dfrac{2}{\sqrt{x}-1}\) cũng nguyên
\(\Rightarrow\left(\sqrt{x}-1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Do: \(\sqrt{x}\ge0\) nên \(\sqrt{x}-1\ge-1\)
\(\Rightarrow\sqrt{x}-1\in\left\{1;-1;2\right\}\)
\(\Rightarrow x\in\left\{4;0;9\right\}\)(đều thỏa mãn ĐKXĐ và x\(\in\)Z )
