a: ĐKXĐ: x>=0; x<>1

b: \(G=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)

c: Khi x=0,16 thì \(G=-0.4\cdot\left(0.4-1\right)=-0.4\cdot\left(-0.6\right)=0.24\)

d: G=-x+căn x

\(=-\left(x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}\right)=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)

Dấu = xảy ra khi x=1/4

\(a,P=\dfrac{-x+2\sqrt{x}-1+x-2\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}:\dfrac{2\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ P=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\\ \Rightarrow P=\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{5-\sqrt{5}}{5}\\ c,\dfrac{P}{\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{1}{\sqrt{x}-1}\le\dfrac{1}{0-1}=-1\)

Vậy \(\left(\dfrac{P}{\sqrt{x}}\right)_{max}=-1\Leftrightarrow x=0\)

a: \(G=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)

b: Khi x=0,16 thì \(G=-0,4\left(0,4-1\right)=-0,4\cdot\left(-0,6\right)=0,24\)

a. ĐKXĐ: x\(\ne1\) x, \(\ne-1\)

b. \(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\)

=\(\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)

=\(\left(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)

\(\left(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\left(\dfrac{\left(\sqrt{x}-2\right)-\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\left(\dfrac{\sqrt{x}-2-\sqrt{x}-2}{\sqrt{x}+1}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\dfrac{4}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1^2}{2}=2\left(\sqrt{x}-1\right)=2\sqrt{x}-2\)

c. khi x=0,16 thì G=\(2\sqrt{x}-2=2\sqrt{0,16}-2=2.0,4-2=0,8-2=-1,2\)

a: ĐKXĐ: x>=0; x<>1

b: \(G=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)^2}{2}\)

\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{2}\cdot\left(\sqrt{x}-1\right)\)

\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)

c: Thay x=0,16 vào G, ta được:

\(H=-0,4\cdot\left(0,4-1\right)=-0,4\cdot0,3=-0,12\)