a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)
b: \(G=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\dfrac{x^2-2x+1}{2}\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{1}\cdot\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\)
c: Khi x=0,16 thì \(G=-\sqrt{0,16}\left(\sqrt{0,16}-1\right)=-0,4\cdot\left(0,4-1\right)=-0,4\cdot\left(-0,6\right)=0,24\)
d: \(G=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}\)
\(=-\left(x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x}-\dfrac{1}{2}=0\)
=>\(x=\dfrac{1}{4}\)
e: G<0
=>\(-\sqrt{x}\left(\sqrt{x}-1\right)< 0\)
=>\(\sqrt{x}\left(\sqrt{x}-1\right)>0\)
=>\(\sqrt{x}-1>0\)
=>\(\sqrt{x}>1\)
=>x>1
h: Khi 0<x<1 thì \(\sqrt{x}-1< 0\)
=>\(\sqrt{x}\left(\sqrt{x}-1\right)< 0\)
=>\(G=-\sqrt{x}\left(\sqrt{x}-1\right)>0\)
