Question

Cho A=(\(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\) ).(\(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\) ) (Với x>0,x\(\ne\)1)

a) Rút gọn.

b)Tính giá trị của x khi A>\(\frac{1}{6}\)

Answer 1

\(A=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)

Dấu giữa 2 biểu thức là dấu chia sẽ hợp lý hơn

Khi đó \(A=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)

\(A>\frac{1}{6}\Rightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}>\frac{1}{6}\Rightarrow2\sqrt{x}-4>\sqrt{x}\)

\(\Rightarrow\sqrt{x}>4\Rightarrow x>16\)