a/ ĐKXĐ:...
\(N=\left(\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right).\frac{4\sqrt{x}}{3}\)
\(N=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{4\sqrt{x}}{3}\)
\(N=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b/ N=\(N=\frac{8}{9}\Rightarrow\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
\(\Leftrightarrow36\sqrt{x}=24x-24\sqrt{x}+24\)
\(\Leftrightarrow24x-60\sqrt{x}+24=0\)
Đặt \(\sqrt{x}=a\ge0\Rightarrow x=a^2\)
\(\Rightarrow24a^2-60a+24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\)
\(a.N=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\\ =\left(\frac{x+2}{\sqrt{x^3}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\\ =\left(\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\cdot\frac{4\sqrt{x}}{3}\\ =\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
\(b.N=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\Leftrightarrow4\sqrt{x}=\frac{8}{3}\left(x-\sqrt{x}+1\right)\\ \Leftrightarrow3\sqrt{x}=2\left(x-\sqrt{x}+1\right)\\ \Leftrightarrow2x-5\sqrt{x}+2=0\\ \left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\)
Bước cuối bạn tự làm nha (do mk bận)
