a) Ta có:
\(x=\sqrt{9+4\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(x=\sqrt{\left(\sqrt{5}\right)^2+2\cdot2\cdot\sqrt{5}+2^2}-\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot1+1^2}\)
\(x=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(x=\left|\sqrt{5}+2\right|-\left|\sqrt{5}+1\right|\)
\(x=\sqrt{5}+2-\sqrt{5}-1\)
\(x=1\)
Thay x=1 vào A ta có:
\(A=\dfrac{\sqrt{1}-1}{\sqrt{1}+2}=\dfrac{1-1}{1+2}=\dfrac{0}{3}=0\)
b) \(A>\dfrac{1}{2}\) khi:
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}>\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{2\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+2\right)}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}>0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-4}{2\left(\sqrt{x}+2\right)}>0\)
Mà: \(2\left(\sqrt{x}+2\right)>0\forall x\)
\(\Leftrightarrow\sqrt{x}-4>0\)
\(\Leftrightarrow x>16\)
Vậy: ....
\(\sqrt{9+4\sqrt{5}}-\sqrt{6-2\sqrt{5}}\\ =\sqrt{5+4\sqrt{5}+4}-\sqrt{5-2\sqrt{5}+1}\\ =\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\\ =\left|\sqrt{5}+2\right|-\left|\sqrt{5}+1\right|\\ =\sqrt{5}+2-\sqrt{5}-1\\ =1\)
`=> x=1`
Thay `x=1` vào `A` ta có :
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{1-1}{1+2}\\ =\dfrac{0}{3}=0\)
__
Để `A>1/2`
Thì : \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}>\dfrac{1}{2}\\ \Leftrightarrow\dfrac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+2\right)}>\dfrac{\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\\ \Leftrightarrow2\sqrt{x}-2>\sqrt{x}+2\\ \Leftrightarrow\sqrt{x}>4\\ \Leftrightarrow x>16\)
\(x=\sqrt{9+4\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}+2-\sqrt{5}-1=1\) ( \(\left\{{}\begin{matrix}\sqrt{5}+2>0\\\sqrt{5}+1>0\end{matrix}\right.\))
a) \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{1-1}{1+2}=0\left(đk:x\ge0\right)\)
b) \(A>\dfrac{1}{2}\)\(\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+2}>\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-2}{2\sqrt{x}+4}>0\)
\(\Leftrightarrow\sqrt{x}>4\) (vì \(2\sqrt{x}+4>0\))
\(\Leftrightarrow x>16\)
