2B:
a) Ta có: \(N=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Ta có: \(N=\dfrac{8}{9}\)
nên \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)
\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)
\(\Leftrightarrow24x-24\sqrt{x}+24-36\sqrt{x}=0\)
\(\Leftrightarrow24x-60\sqrt{x}+24=0\)
\(\Leftrightarrow2x-5\sqrt{x}+2=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=\dfrac{1}{4}\left(nhận\right)\end{matrix}\right.\)
2A:
a) Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{2\sqrt{x}+2-2+x}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)}\cdot\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x}{\sqrt{x}-1}\)
b) Để \(N=\dfrac{8}{9}\) thì \(9x=8\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow9x-8\sqrt{x}+8=0\)
\(\Leftrightarrow x-\dfrac{8}{9}\sqrt{x}+\dfrac{8}{9}=0\)
\(\Leftrightarrow x-2\cdot\sqrt{x}\cdot\dfrac{4}{9}+\dfrac{16}{81}+\dfrac{56}{81}=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{4}{9}\right)^2+\dfrac{56}{81}=0\)(Vô lý)
Vậy: Không có giá trị nào của x để \(N=\dfrac{8}{9}\)
