\(a,\dfrac{x+2}{x\sqrt{x}-1}=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(b,c,B=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\left(x\ge0;x\ne1\right)\\ B=\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ B=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ B=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(d,x=8\Leftrightarrow B=\dfrac{2\sqrt{2}}{8+2\sqrt{2}+1}=\dfrac{2\sqrt{2}}{9+2\sqrt{2}}=\dfrac{2\sqrt{2}\left(9-2\sqrt{2}\right)}{73}=\dfrac{18\sqrt{2}-8}{73}\)
\(f,B=1\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=1\\ \Leftrightarrow\sqrt{x}=x+\sqrt{x}+1\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\left(L\right)\Leftrightarrow x\in\varnothing\)
\(g,B=3\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=3\\ \Leftrightarrow3x+3\sqrt{x}+3=\sqrt{x}\\ \Leftrightarrow3x+2\sqrt{x}+3=0\\ \Leftrightarrow x\in\varnothing\left(3x+2\sqrt{x}+3>0\right)\)
\(h,B>4\Leftrightarrow B-4>0\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-4>0\\ \Leftrightarrow\dfrac{\sqrt{x}-4x-4\sqrt{x}-4}{x+\sqrt{x}+1}>0\\ \Leftrightarrow\dfrac{-4x-3\sqrt{x}-4}{x+\sqrt{x}+1}>0\\ \Leftrightarrow-4x-3\sqrt{x}-4>0\left(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\right)\\ \Leftrightarrow x\in\varnothing\left(-4x-3\sqrt{x}-4=-\left(4x+3\sqrt{x}+4\right)< 0\right)\)
\(i,B=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}+1+\sqrt{x}\\ \ge2\sqrt{\dfrac{1}{\sqrt{x}}\cdot\sqrt{x}}+1=2\cdot1+1=3\left(BĐTcosi\right)\)
Dấu \("="\Leftrightarrow\dfrac{1}{\sqrt{x}}=\sqrt{x}\Leftrightarrow x=1\)
g: Để B=3 thì \(\sqrt{x}=3x+3\sqrt{x}+3\)
\(\Leftrightarrow3x+2\sqrt{x}+3=0\)
\(\Leftrightarrow x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{8}{9}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{8}{9}=0\)(vô lý)
