Question

Cho biểu thức: P = \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right)\); Q = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-1\right)\) với x ≥ 0; x ≠ 1

a, Rút gọn biểu thức M = P : Q

b, Tìm x để M <\(\dfrac{3}{2}\)

Answer 1

a: \(M=P:Q\)

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

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

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

b: M<3/2

=>\(M-\dfrac{3}{2}< 0\)

=>\(\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}-\dfrac{3}{2}< 0\)

=>\(\dfrac{2\left(2\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}+1\right)}< 0\)

=>\(2\left(2\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)< 0\)

=>\(\sqrt{x}-1< 0\)

=>\(\sqrt{x}< 1\)

=>0<=x<1