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ĐKXĐ: \(x\ge0;x\ne1\)
\(G=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\frac{\left(x-1\right)^2}{2}\)
\(=\left(\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right)\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}.\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}=-\sqrt{x}\left(\sqrt{x}-1\right)=\sqrt{x}-x\)
\(x=0,16\Rightarrow G=\sqrt{0,16}-0,16=\)
\(G=\frac{1}{4}-x+\sqrt{x}-\frac{1}{4}=\frac{1}{4}-\left(\sqrt{x}-\frac{1}{2}\right)^2\le\frac{1}{4}\)
\(\Rightarrow G_{max}=\frac{1}{4}\) khi \(\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)
G nguyên khi \(\sqrt{x}\) nguyên \(\Rightarrow x=k^2\) với \(k\in Z\)
Vậy với mọi x có dạng \(x=k^2\) thì G nguyên
\(G=\sqrt{x}\left(1-\sqrt{x}\right)\)
Với \(0< x< 1\Rightarrow0< \sqrt{x}< 1\Rightarrow1-\sqrt{x}>0\)
\(\Rightarrow G=\sqrt{x}\left(1-\sqrt{x}\right)>0\Rightarrow G\) dương
Để \(G< 0\Rightarrow1-\sqrt{x}< 0\Rightarrow x>1\)
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=\(2\sqrt{x}-2=2\sqrt{0,16}-2=2.0,4-2=0,8-2=-1,2\)