\(\frac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\)

\(=\frac{1}{\sqrt{n-1}\sqrt{n}\left(\sqrt{n-1}+\sqrt{n}\right)}\)

\(=\frac{\sqrt{n-1}-\sqrt{n}}{\sqrt{n-1}\sqrt{n}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n-1}}\)

Sau đó bạn tự áp dụng vào nhé!

\(\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)

\(=\dfrac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\)

\(=\dfrac{\sqrt{3}+\sqrt{2}}{3-2}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}-\sqrt{3}\)

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

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

\(=\sqrt{3}\)

\(\sqrt{3}-\frac{5}{2}>\sqrt{3}-4\text{ vì }-\frac{5}{2}>-4\)

\(\Rightarrow2.\left(\sqrt{3}-\frac{5}{2}\right)>\sqrt{3}-4\)

\(\Rightarrow2.\sqrt{3}-5>\sqrt{3}-4\)

b) vì \(\sqrt{5}-\sqrt{12}< 0\), ta có: 

 \(5\sqrt{5}-2\sqrt{3}=4\sqrt{5}+\sqrt{5}-\sqrt{12}< 4\sqrt{5}< 4\sqrt{5}+6\) 

Vậy \(5\sqrt{5}-2\sqrt{3}< 6+4\sqrt{5}\)

c)\(\sqrt{2}-\sqrt{6}=\sqrt{2}.\left(\sqrt{1}-\sqrt{3}\right)>\left(1-\sqrt{3}\right)\)

Vậy \(\sqrt{2}-\sqrt{6}>1-\sqrt{3}\)

a,Ta có :  \(1-\sqrt{3}\); \(\sqrt{2}-\sqrt{6}=\sqrt{2}\left(1-\sqrt{3}\right)\Rightarrow1-\sqrt{3}< \sqrt{2}\left(1-\sqrt{3}\right)\)

Vậy \(1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

b, Đặt A =  \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)(*)

\(\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{7}+1-\sqrt{7}+1-2=0\Rightarrow A=0\)

Vậy (*) = 0 

1: 

Ta có: \(\sqrt{2}-\sqrt{6}\)

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

\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

2:
Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)

\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}\)

=0