Question

rút gọn biểu thức

a) \(5\sqrt{\frac{1}{5}}+\frac{1}{3}\sqrt{45}+\sqrt{\left(2-\sqrt{5}\right)^2}\)

b) \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)

cho biểu thức

A= \(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right).\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\) với x>0; x khác 0

a) rút gọn biểu thức A

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

Answer 1

1: Rút gọn biểu thức

a) Ta có: \(5\sqrt{\frac{1}{5}}+\frac{1}{3}\sqrt{45}+\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(=5\cdot\frac{1}{\sqrt{5}}+\frac{1}{3}\cdot3\sqrt{5}+\left|2-\sqrt{5}\right|\)

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-2\)(Vì \(2< \sqrt{5}\))

\(=3\sqrt{5}-2\)

b) Ta có: \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)

\(=\frac{\left(5+\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\frac{\left(5-\sqrt{5}\right)^2}{\left(5+\sqrt{5}\right)\left(5-\sqrt{5}\right)}\)

\(=\frac{30+10\sqrt{5}+30-10\sqrt{5}}{25-5}\)

\(=\frac{60}{20}=3\)

2:

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

a) Ta có: \(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

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

\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

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

\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)

b) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4;\frac{14\pm6\sqrt{5}}{4}\right\}\end{matrix}\right.\)

Để \(A>\frac{1}{6}\) thì \(A-\frac{1}{6}>0\)

\(\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}-\frac{1}{6}>0\)

\(\Leftrightarrow\frac{2\sqrt{x}-4}{6\sqrt{x}}-\frac{\sqrt{x}}{6\sqrt{x}}>0\)

\(\Leftrightarrow\frac{\sqrt{x}-4}{6\sqrt{x}}>0\)

mà \(6\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}-4>0\)

\(\Leftrightarrow\sqrt{x}>4\)

hay x>16

Kết hợp ĐKXĐ, ta được: x>16

Vậy: Để \(A>\frac{1}{6}\)thì x>16