Đáp án:
a)Đk: x>0; \(x\neq1\)
A=\(\frac{\sqrt{x}}{\sqrt{x}+1}\)
B=\(\frac{\sqrt{x}-2}{\sqrt{x}}\)
b)\(x\geq 4\)
x< 4
Giải thích các bước giải:
Đk: x>0; \(x\neq1\)
A=\(\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
B=\(\frac{x-4}{x+2\sqrt{x}}=\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+2)}=\frac{\sqrt{x}-2}{\sqrt{x}}\)
b)Nếu A≥B
⇒\(\frac{\sqrt{x}}{\sqrt{x}+1}\geq \frac{\sqrt{x}-2}{\sqrt{x}}\)
⇔\(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-2}{\sqrt{x}}\geq 0\)
⇔\(\frac{x-x+\sqrt{x}+2}{\sqrt{x}(\sqrt{x}+1)}\geq 0\)
⇔\(\frac{\sqrt{x}+2}{\sqrt{x}(\sqrt{x}+1)}\geq 0\)
⇔\(\sqrt{x}+2\geq 0 ⇔\sqrt{x}\geq -2⇔ x\geq 4\)
Nếu A>B
⇒\(\frac{\sqrt{x}}{\sqrt{x}+1}< \frac{\sqrt{x}-2}{\sqrt{x}}\)
⇔\(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-2}{\sqrt{x}}<0\)
⇔\(\frac{x-x+\sqrt{x}+2}{\sqrt{x}(\sqrt{x}+1)}< 0\)
⇔\(\frac{\sqrt{x}+2}{\sqrt{x}(\sqrt{x}+1)}<0\)
⇔\(\sqrt{x}+2<0 ⇔\sqrt{x}<-2⇔ x< 4\)
