Với n > 0 Ta có:

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

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

\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)

\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)

\(\sqrt{16}+\sqrt{9}=3+4=7\)

a: \(\sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2}\)

b: \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)

c: \(\sqrt{16+6\sqrt{7}}=4+\sqrt{7}\)

d: \(\sqrt{31-12\sqrt{3}}=3\sqrt{3}-2\)

e: \(\sqrt{27+10\sqrt{2}}=5+\sqrt{2}\)

f: \(\sqrt{14+6\sqrt{5}}=3+\sqrt{5}\)

1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)

 

Cho mình sửa đề xí ạ! 

b) \(\frac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}\)

\(\sqrt{7+2\sqrt{3}}-\sqrt{7-2\sqrt{3}};\left(\sqrt{7+\sqrt{12}}-\sqrt{7-\sqrt{12}}\right)^2=7+\sqrt{12}-\sqrt{12}+7-2\sqrt{\left(7+\sqrt{12}\right)\left(7-\sqrt{12}\right)}=14-2\sqrt{37}\Rightarrow\sqrt{7+\sqrt{12}}-\sqrt{7-\sqrt{12}}=\sqrt{14-2\sqrt{37}}\)

a) \(\sqrt{10+2\sqrt{14}}\cdot\sqrt{10+2\sqrt{14}}\)

\(=\sqrt{\left(10+2\sqrt{14}\right)^2}\)

\(=10+2\sqrt{14}\)

b) \(\sqrt{7+\sqrt{12}}-\sqrt{7-\sqrt{12}}\)

\(=\sqrt{\left(\sqrt{7+\sqrt{12}}-\sqrt{7-\sqrt{12}}\right)^2}\)

\(=\sqrt{7+\sqrt{12}+7-\sqrt{12}-2\sqrt{\left(7+\sqrt{12}\right)\left(7-\sqrt{12}\right)}}\)

\(=\sqrt{14-2\sqrt{49-12}}\)

\(=\sqrt{14-2\sqrt{37}}\)

\(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\)

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

\(=\dfrac{\sqrt{5}}{\sqrt{4}}\)

\(=\dfrac{\sqrt{5}}{2}\)

\(\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}\)

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

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

\(=-\dfrac{\sqrt{3}}{\sqrt{7}}\)

\(=-\dfrac{\sqrt{21}}{7}\)

____________

\(\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\)

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

\(=\dfrac{\sqrt{5}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{10}}{2}\)

\(a,\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\\ =\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{\sqrt{2}.\sqrt{4}-\sqrt{3}.\sqrt{4}}\\ =\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2^2}}\\ =\dfrac{\sqrt{5}}{2}\)

\(b,\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\\ =\dfrac{\sqrt{5}.\sqrt{3}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}\)

\(c,\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\dfrac{\sqrt{5}}{\sqrt{2}}\)

\(a,=\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}\\ =\dfrac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{2}\)

\(b,=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}=\dfrac{\sqrt{21}}{7}\)

\(c,=\dfrac{\sqrt{5}.\sqrt{5}+\sqrt{5}}{\sqrt{2}.\sqrt{5}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{10}}{2}\)

Bạn xem lại câu 5 xem có sai đề không chứ mình tính mãi không ra

1. \(\sqrt{14+2\sqrt{33}}=\sqrt{\left(\sqrt{11}+\sqrt{3}\right)^2}=\sqrt{11}+\sqrt{3}\)

2. \(\sqrt{29-12\sqrt{5}}=\sqrt{\left(\sqrt{20}-\sqrt{9}\right)^2}=2\sqrt{5}-3=-3+2\sqrt{5}\)

3. \(\sqrt{16+2\sqrt{55}}=\sqrt{\left(\sqrt{11}+\sqrt{5}\right)^2}=\sqrt{11}+\sqrt{5}\)

4. \(\sqrt{13+4\sqrt{10}}=\sqrt{\left(\sqrt{8}+\sqrt{5}\right)^2}=2\sqrt{2}+\sqrt{5}\)

5. \(\sqrt{36+12\sqrt{5}}=\sqrt{\left(\sqrt{30}+\sqrt{6}\right)^2}=\sqrt{30}+\sqrt{6}\)

6. \(\sqrt{21-6\sqrt{6}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}=3\sqrt{2}-\sqrt{3}=-\sqrt{3}+3\sqrt{2}\)

P/s: Đây là dạng toán căn lồng căn, được dùng máy tính để biến đổi thành căn bình phương, nếu bạn chưa biết thì search gg nhé.

Mình cần trình bày ạ!