Question

Tìm `x` để `AB. B > 3/2`, biết:
\(A=\dfrac{3+\sqrt{x}}{\sqrt{x}}\)

\(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

Answer 1

Tìm \(x\) để \(A.B>\dfrac{3}{2}?\)

 \(ĐK:x>0;\\ A.B\\ =\dfrac{3+\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =1+\dfrac{1}{\sqrt{x}}\\ A.B>\dfrac{3}{2}\\ \Leftrightarrow1+\dfrac{1}{\sqrt{x}}>\dfrac{3}{2}\\ \Leftrightarrow\dfrac{1}{\sqrt{x}}>\dfrac{3}{2}-1\\ \Leftrightarrow\dfrac{1}{\sqrt{x}}>\dfrac{1}{2}\\ \Leftrightarrow\sqrt{x}< 2\\ \Leftrightarrow0< x< 4\)

Vậy \(0< x< 4\) thì \(A.B>\dfrac{3}{2}\)