`0<`$\dfrac{{2}}{x+\sqrt{x}+1}$`<=2`
ta có:$\dfrac{{2}}{x+\sqrt{x}+1}$
`=`$\dfrac{{2}}{x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}4}$
`=`$\dfrac{{2}}{(\sqrt{x}+\dfrac{1}2)^{2}+\dfrac{3}4}$
vì `(`$\sqrt[]{x}+$`1/2)^2>=0;3/4>0(∀x)`
`=>` $\dfrac{{2}}{x+\sqrt{x}+1}$`>0`
lại có $x+\sqrt[]{x}+1>0$ mà `2>0`
`=>`$\dfrac{{2}}{x+\sqrt{x}+1}$`>=2`
dấu`'='` sẩy ra `<=>``x+` $\sqrt[]{x}+1$`=1`
`<=> x=0`
