Lời giải:

Gọi tổng trên là $A$. Ta có:

\(2A>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)

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

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

\(2A>\sqrt{81}-1=8\Rightarrow A>4\)

Ta có đpcm.

\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}...+\dfrac{1}{\sqrt{79}+\sqrt{80}}\)

\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+...+\left(\sqrt{80}-\sqrt{79}\right)\)

\(=\sqrt{80}-\sqrt{2}\)

Đến đây bấm máy rồi đối chiếu kết quả cho nhanh, hoặc nếu em thik "màu mè" hơn thì giả sử lớn hơn rồi biến đổi tương đương thôi :)

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)

Lời giải:

\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)

\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+...+\frac{2}{\sqrt{79}+\sqrt{80}}\)

\(>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+....+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)

\(2A>\frac{(\sqrt{2}-\sqrt{1})(\sqrt{2}+\sqrt{1})}{\sqrt{1}+\sqrt{2}}+\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}{\sqrt{2}+\sqrt{3}}+....+\frac{(\sqrt{80}-\sqrt{79})(\sqrt{80}+\sqrt{79})}{\sqrt{79}+\sqrt{80}}+\frac{(\sqrt{81}-\sqrt{80})(\sqrt{81}+\sqrt{80})}{\sqrt{80}+\sqrt{81}}\)

\(2A>(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+...+(\sqrt{80}-\sqrt{79})+(\sqrt{81}-\sqrt{80})\)

\(2A>\sqrt{81}-\sqrt{1}=8\)

\(A>4\) (đpcm)

Lời giải:
a.

\(=2\sqrt{4^2.5}+3\sqrt{3^2.5}-\sqrt{7^2.5}=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}=10\sqrt{5}\)

b.

\(=\frac{3(2-\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{13(4+\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})}+\frac{6\sqrt{3}}{3}\)

\(=\frac{6-3\sqrt{3}}{1}+\frac{13(4+\sqrt{3})}{13}+2\sqrt{3}=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)

\(=10\)

c.

\(=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}\right].(\sqrt{7}-\sqrt{5})\)

\(=(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=7-5=2\)

d.

\(=|2+\sqrt{3}|-\sqrt{5^2-2.5\sqrt{3}+3}=|2+\sqrt{3}|-\sqrt{(5-\sqrt{3})^2}\)

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

a) Ta có: \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)

\(=10\sqrt{5}\)

b) Ta có: \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=3\left(2-\sqrt{3}\right)+4+\sqrt{3}+2\sqrt{3}\)

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

\(=10+\sqrt{3}\)

c) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

=7-5=2

d) Ta có: \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)

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

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

a. \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)

\(=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)

\(=10\sqrt{5}\)

b. \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\dfrac{6\sqrt{3}}{\sqrt{3}.\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{4-3}+\dfrac{13\left(4+\sqrt{3}\right)}{16-3}+\dfrac{6\sqrt{3}}{3}\)

\(=3\left(2-\sqrt{3}\right)+\dfrac{13\left(4+\sqrt{3}\right)}{13}+2\sqrt{3}\)

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

\(=10\)

c. \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=7-5=2\)

d. \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)

\(=\left|2+\sqrt{3}\right|-\sqrt{5^2-2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

\(=\left|2+\sqrt{3}\right|-\left|5-\sqrt{3}\right|\)

\(=2+\sqrt{3}-\left(5-\sqrt{3}\right)\) (vì \(\left|2+\sqrt{3}\right|\ge0,\left|5-\sqrt{3}\right|\ge0\))

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

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

Bài 1: 

a) \(\sqrt{1,44\cdot1,21-1,44\cdot0,4}\)

\(=\sqrt{1,44\cdot\left(1,21-0,4\right)}\)

\(=\sqrt{1,44\cdot0,81}\)

\(=\sqrt{1,44}\cdot\sqrt{0,81}\)

\(=1,2\cdot0,9\)

\(=1,08\)

b) \(\dfrac{\sqrt{5}-2}{\sqrt{5}+2}+\sqrt{80}\)

\(=\dfrac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+4\sqrt{5}\)

\(=\dfrac{5-4\sqrt{5}+4}{1}+4\sqrt{5}\)

\(=9-4\sqrt{5}+4\sqrt{5}\)

\(=9\)

c) \(\sqrt[3]{16}+\sqrt[3]{2}\left(\sqrt[3]{4}-\sqrt[3]{2}\right)\)

\(=\sqrt[3]{2^3\cdot2}+\sqrt[3]{2\cdot4}-\sqrt[3]{2\cdot2}\)

\(=2\sqrt[3]{2}+\sqrt[3]{8}-\sqrt[3]{4}\)

\(=2\sqrt[3]{2}+2-\sqrt[3]{4}\)

Bài 2: Ta có: 

\(VT=\dfrac{1}{\sqrt{a}-\sqrt{b}}:\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)

\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}:\dfrac{\sqrt{ab}\cdot\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\cdot\dfrac{1}{\sqrt{a}+\sqrt{b}}\)

\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{1}{a-b}=VP\left(dpcm\right)\)