Thay x=0,25 vào A ta có:
\(A=\dfrac{2+\sqrt{0,25}}{\sqrt{0,25}}\)
A = 5
Vậy với x=0,25 thì A =5
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}} +\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\) (ĐK: x\(\ge0\))
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x+\sqrt{x}-\sqrt{x}-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x} +2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\) (ĐPCM)
c) \(\dfrac{A}{B}\ge\dfrac{3}{2}\)
<=> \(\dfrac{2+\sqrt{x}}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)\(\ge\dfrac{3}{2}\)
<=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}\ge\dfrac{3}{2}\)
<=> \(3\sqrt{x}=2\sqrt{x}+2\) (Vì x \(\ge0\) )
<=>\(\sqrt{x}=2\)
<=> \(x=4\)
Vậy với x = 4 thì thỏa mãn yêu cầu bài toán.
P/s: Lần sau đăng tách từng câu hỏi nhỏ ra nha bạn =))
