\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)

\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)

\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)

\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)

\(=\frac{1-\sqrt{25}}{-1}=4\)

\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)

\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)

\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)

\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)

\(=1\)

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

=>K=1-\( \frac{1}{5} \)=\(\frac{4}{5} \)

Câu a, b, bạn có thể làm được suy nghĩ đi nha

c)

Ta có pt tổng quát :

\(\dfrac{1}{a\sqrt{a+1}+\left(a+1\right)\sqrt{a}}=\dfrac{1}{\sqrt{a\left(a+1\right)}\left(\sqrt{a}+\sqrt{\left(a+1\right)}\right)}=\dfrac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a}\sqrt{a+1}}=\dfrac{1}{\sqrt{a}}-\dfrac{1}{\sqrt{a+1}}\)\(\Rightarrow C=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+.....+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}=1-\dfrac{1}{5}=\dfrac{4}{5}\)..........Kaito Kid.......

a)=-14

C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)

C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\)

C = \(\dfrac{2\sqrt{4-\sqrt{6+\sqrt{20}}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}\)

C = \(\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}\) = \(\dfrac{2\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)}{10-2}\)

C = \(\dfrac{2\sqrt{30-10\sqrt{5}}+2\sqrt{6-2\sqrt{5}}}{8}\)

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

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

C = \(\dfrac{10-2\sqrt{5}+2\sqrt{5}-2}{8}\) = \(\dfrac{8}{8}\) = \(1\)

D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)

D = \(\sqrt{\left(7-3\sqrt{5}\right)^2}-\sqrt{\left(7+3\sqrt{5}\right)^2}\)

D = \(7-3\sqrt{5}-\left(7+3\sqrt{5}\right)\) = \(7-3\sqrt{5}-7-3\sqrt{5}\)

D = \(-6\sqrt{5}\)

A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

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

A = \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\) = \(\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

A = \(\sqrt{\sqrt{5}-\sqrt{5}+1}\) = \(\sqrt{1}=1\)

\(A=-\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+....-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}\)

\(A=\sqrt{9}-\sqrt{1}=3-1=2\)

b) bạn trục mẫu đi nha dựa vào hằng đẳng thức a^2 -b^2=(a-b)(a+b)

rồi bạn tính nói chung mẫu bằng -1

tính cái trên tử kết quả là 4

c) bạn dựa vào câu b .\(\dfrac{1}{\sqrt{3}}=\dfrac{2}{2\sqrt{3}}>\dfrac{2}{\sqrt{3}+\sqrt{4}}\)

từ đó suy ra B > 2A vậy B>8

\(A=\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)

\(=\dfrac{\sqrt{7-2\sqrt{6}}-1}{7-2\sqrt{6}-1}-\dfrac{\sqrt{7+2\sqrt{6}}-1}{7+2\sqrt{6}-1}\)

\(=\dfrac{\sqrt{\left(\sqrt{6}-1\right)^2}-1}{6-2\sqrt{6}}-\dfrac{\sqrt{\left(\sqrt{6}+1\right)^2}-1}{6+2\sqrt{6}}\)

\(=\dfrac{\sqrt{6}-2}{\sqrt{6}\left(\sqrt{6}-2\right)}-\dfrac{\sqrt{6}}{\sqrt{6}\left(\sqrt{6}+2\right)}\)

\(=\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{6}+2}=\dfrac{\sqrt{6}+2-\sqrt{6}}{\sqrt{6}\left(\sqrt{6}+2\right)}\)

\(=\dfrac{2}{\sqrt{12}\left(\sqrt{3}+\sqrt{2}\right)}=\dfrac{2\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{3}\left(3-2\right)}=\dfrac{3-\sqrt{6}}{3}\)

\(5-2\sqrt{6}=\left(\sqrt{2}\right)^2-2\times\sqrt{2}\times\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{3}-\sqrt{2}\right)^2\)

\(7+2\sqrt{10}=\left(\sqrt{2}\right)^2+2\times\sqrt{2}\times\sqrt{5}+\left(\sqrt{5}\right)^2=\left(\sqrt{2}+\sqrt{5}\right)^2\)

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

\(B=\dfrac{2}{\sqrt{8-2\sqrt{15}}}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}-\dfrac{3}{\sqrt{7+2\sqrt{10}}}\)

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

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

\(=\sqrt{5}+\sqrt{3}-\sqrt{3}-\sqrt{2}-\sqrt{5}+\sqrt{2}=0\)

Câu b nhé:

Ta có:

\(\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+\dfrac{1}{\sqrt{23}+\sqrt{22}}+...+\dfrac{1}{\sqrt{2}+\sqrt{1}}\\ =\dfrac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\dfrac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}+\sqrt{1}\right)\left(\sqrt{2}-\sqrt{1}\right)}\\ =\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}\\ =5-1=4\left(đpcm\right)\)

a) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}=0\) (*)

\(\Leftrightarrow\left(3\sqrt{2}-\sqrt{3}\right)+\left(\sqrt{3}+\sqrt{6}\right)-\left(3+\sqrt{3}\right)\cdot\sqrt{2}=0\)

\(\Leftrightarrow0=0\) (luôn đúng)

Vậy (*) luôn đúng

bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không