a, A = \(\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\) (x > 0; x \(\ne\) 4)
A = \(\dfrac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
A = \(\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3}\)
A = \(\dfrac{2\left(\sqrt{x}+1\right)}{3\sqrt{x}}\)
A = \(\dfrac{2\sqrt{x}+2}{3\sqrt{x}}\)
Vậy ...
b, Ta có: B = A\(\sqrt{x}\) - x = \(\dfrac{2\sqrt{x}+2}{3\sqrt{x}}\cdot\sqrt{x}-x\) = \(\dfrac{2x+2\sqrt{x}}{3\sqrt{x}}-\dfrac{3x\sqrt{x}}{3\sqrt{x}}\) = \(\dfrac{2x-3x\sqrt{x}+2\sqrt{x}}{3\sqrt{x}}\) = \(\dfrac{2\sqrt{x}-3x+2}{3}\) = \(\dfrac{\sqrt{x}\left(2-3\sqrt{x}\right)+2}{3}\)
Ta có: x > 0 \(\Rightarrow\) \(\sqrt{x}>0\)
\(\Rightarrow\) Để B đạt giá trị lớn nhất thì 2 - 3\(\sqrt{x}\) \(\le\) 0
Dấu "=" xảy ra \(\Leftrightarrow\) 2 - 3\(\sqrt{x}\) = 0 \(\Leftrightarrow\) 3\(\sqrt{x}\) = 2 \(\Leftrightarrow\) \(\sqrt{x}\) = \(\dfrac{2}{3}\) \(\Leftrightarrow\) x = \(\dfrac{4}{9}\) (TM)
Khi đó MaxB = \(\dfrac{2}{3}\)
Vậy B đạt GTLN là \(\dfrac{2}{3}\) khi và chỉ khi x = \(\dfrac{4}{9}\)
Chúc bn học tốt!
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)
\(=\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3}\)
\(=\dfrac{2\sqrt{x}+2}{3\sqrt{x}}\)
