\(\dfrac{4}{\sqrt{5}-1}+\dfrac{3}{\sqrt{5}-2}-\dfrac{16}{3-\sqrt{5}}\)

\(=\sqrt{5}+1+3\sqrt{5}+6-12-4\sqrt{5}\)

=-5

Lời giải:

Đặt biểu thức đã cho là $A$

Ta có:

\(\frac{1}{1+\sqrt{2}}+\frac{1}{1+\sqrt{2}}> \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\)

\(\Rightarrow \frac{1}{1+\sqrt{2}}> \frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)

Hoàn toàn TT: \(\frac{1}{\sqrt{3}+\sqrt{4}}> \frac{1}{2}\left(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\right)\)

.......

\(\frac{1}{\sqrt{79}+\sqrt{80}}> \frac{1}{2}\left(\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)

Cộng các bđt trên lại với nhau:

\(\Rightarrow A> \frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)

\(A> \frac{1}{2}\left(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{81}-\sqrt{80}}{81-80}\right)\) (liên hợp)

\(A> \frac{1}{2}> (\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{81}-\sqrt{80})\)

\(A> \frac{1}{2}(\sqrt{81}-1)=4\) (đpcm)

Đề bài đúng: \(\dfrac{\sqrt{4-\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{2}}=1\)

Hoặc: \(\dfrac{\sqrt{4+\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}}=1\)

\(=\dfrac{\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)}{2}=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}=\dfrac{5-3}{2}=1\)

\(\dfrac{\sqrt{4-\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{2}}=\dfrac{\sqrt{8-2\sqrt{15}}\left(\sqrt{5}+\sqrt{3}\right)}{2}\)

\(=\dfrac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\left(\sqrt{5}+\sqrt{3}\right)}{2}=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{2}=\dfrac{5-3}{2}=1\)

help

  => đề sai rồi bạn

\(a,=6\sqrt{3}-10\sqrt{3}+15\sqrt{3}=11\sqrt{3}\\ b,=2\sqrt{5}-\sqrt{5}+70\sqrt{5}=71\sqrt{5}\\ c,=\dfrac{\sin43^0}{\sin43^0}+1=1+1=2\\ d,Sửa:\dfrac{\tan32^0}{\cot68^0}-\cos30^0-\dfrac{\sin18^0}{\sin82^0}=\dfrac{\tan32^0}{\tan32^0}-\dfrac{\sqrt{3}}{2}-\dfrac{\sin18^0}{\cos18^0}=1-1-\dfrac{\sqrt{3}}{2}=-\dfrac{\sqrt{3}}{2}\)

DONE!

\(\dfrac{\sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6}+\sqrt{8}+\sqrt{10}}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)

\(=\dfrac{\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)+\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{4}+\sqrt{2}.\sqrt{5}}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)

\(=\dfrac{\left(\sqrt{3}+\sqrt{4}+\sqrt{5}\right)\left(1+\sqrt{2}\right)}{\sqrt{3}+\sqrt{4}+\sqrt{5}}\)

\(=1+\sqrt{2}\)

⇒ ĐPCM

\(=\left(\dfrac{2}{5}+\dfrac{2}{5}\sqrt{6}+3+\dfrac{3}{2}\sqrt{6}-3-\sqrt{6}\right)\cdot\dfrac{5}{9\sqrt{6}+4}\)

\(=\left(\dfrac{2}{5}+\dfrac{9}{10}\sqrt{6}\right)\cdot\dfrac{5}{9\sqrt{6}+4}\)

\(=\dfrac{5}{10}=\dfrac{1}{2}\)

Đặt A=\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}\)

\(\Leftrightarrow A=\dfrac{2}{2\sqrt{2}}+\dfrac{2}{2\sqrt{3}}+....+\dfrac{2}{2\sqrt{100}}\)

\(\Leftrightarrow A=\dfrac{2}{\sqrt{2}+\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{3}}+....+\dfrac{2}{\sqrt{99}+\sqrt{99}}+\dfrac{2}{\sqrt{100}+\sqrt{100}}\)

\(\Leftrightarrow A=2\left(\dfrac{1}{\sqrt{2}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{99}}+\dfrac{1}{\sqrt{100}+\sqrt{100}}\right)\)

Ta có:

\(\dfrac{1}{\sqrt{2}+\sqrt{2}}< \dfrac{1}{1+\sqrt{2}};\dfrac{1}{\sqrt{3}+\sqrt{3}}< \dfrac{1}{\sqrt{2}+\sqrt{3}}\)

Tường tự, ta có:

\(\dfrac{A}{2}< \dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(A< 2\left(\dfrac{1-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{99}-\sqrt{100}}{-1}\right)\)

\(A< -2\left(1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...-\sqrt{99}+\sqrt{99}-\sqrt{100}\right)\)

\(A< -2\left(1-\sqrt{100}\right)\)

\(A< 18\)

Vậy\(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{100}}< 18\)