Bài 7:
a: Ta có: \(C=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
\(7,\\ a,C=\dfrac{1}{\sqrt{x}-1}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ C=\dfrac{x\sqrt{x}+1-3\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\\ C=\dfrac{x\sqrt{x}+1-3\sqrt{x}+3+2x-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\\ C=\dfrac{2x+x\sqrt{x}-3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(8,\\ a,C=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\left(x\ge0;x\ne1\right)\\ C=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ C=\dfrac{-2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-1}{\sqrt{x}+1}\\ b,x=\dfrac{4}{9}\Leftrightarrow C=\dfrac{-1}{\dfrac{2}{3}+1}=\dfrac{-1}{\dfrac{5}{3}}=-\dfrac{3}{5}\)
