Question

 Với x > 0, cho hai biểu thức A = \(\dfrac{x+2\sqrt{x}}{x}\) và  B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)

a, Chứng minh B = \(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

b, Tìm số nguyên x nhỏ nhất để \(\dfrac{A}{B}\) < \(\dfrac{7}{4}\)

Answer 1

a: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

b: \(\dfrac{A}{B}< \dfrac{7}{4}\)

=>\(\dfrac{x+2\sqrt{x}}{x}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}< \dfrac{7}{4}\)

=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{7}{4}< 0\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{7}{4}< 0\)

=>\(\dfrac{4\left(\sqrt{x}+1\right)-7\sqrt{x}}{4\sqrt{x}}< 0\)

=>\(\dfrac{-3\sqrt{x}+4}{4\sqrt{x}}< 0\)

=>\(-3\sqrt{x}+4< 0\)

=>\(-3\sqrt{x}< -4\)

=>\(\sqrt{x}>\dfrac{4}{3}\)

=>\(x>\dfrac{16}{9}\)

=>Số nguyên x nhỏ nhất thỏa mãn là x=1