\(B=\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}.\)

\(=\frac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}+...+\frac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{99}+\sqrt{97}\right)\left(\sqrt{99}-\sqrt{97}\right)}.\)

\(=\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{99}-\sqrt{97}}{99-97}.\)

\(=\frac{\sqrt{5}}{2}-\frac{\sqrt{3}}{2}+\frac{\sqrt{7}}{2}-\frac{\sqrt{5}}{2}+...+\frac{\sqrt{99}}{2}-\frac{\sqrt{97}}{2}=\frac{\sqrt{99}}{2}-\frac{\sqrt{3}}{2}\)

Vậy \(B=\frac{\sqrt{99}-\sqrt{3}}{2}.\)

bài này dễ mà bạn cần mk giải chi tiết ko

kết quảA =\(\frac{\sqrt{99}-\sqrt{3}}{2}\)

Có: \(\frac{1}{\sqrt{n}+\sqrt{n+2}}=\frac{\sqrt{n+2}-\sqrt{n}}{\left(\sqrt{n+2}-\sqrt{n}\right)\left(\sqrt{n+2}+\sqrt{n}\right)}=\frac{\sqrt{n+2}-\sqrt{n}}{2}\)
\(\Rightarrow A=\frac{\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{99}-\sqrt{97}}{2}\)
\(A=\frac{\sqrt{99}-\sqrt{3}}{2}\)

\(2A=\frac{2}{1+\sqrt{3}}+\frac{2}{\sqrt{5}+\sqrt{7}}+...+\frac{2}{\sqrt{97}+\sqrt{99}}\)

\(2A>\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}+\frac{1}{\sqrt{99}+\sqrt{101}}\)

Nhân liên hợp tử - mẫu vế phải:

\(\Rightarrow2A>\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)

\(2A>\frac{1}{2}\left(\sqrt{101}-1\right)>\frac{1}{2}\left(\sqrt{100}-1\right)=\frac{9}{2}\)

\(\Rightarrow A>\frac{9}{4}\)

với n >0, ta có :

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

Gọi biểu thức đã cho là A

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

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

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

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

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

Bạn trục căn thức ở mẫu rồi trừ đi là xong nhé,vì khi trục căn thức thì ở A mẫu chung là 1,ở B mẫu chung là 2.

giai ra giup mik di

A=(√3-√2)/(3-2)+(√4-√3)/(4-3)+......

=√3-√2+√4-√3+......+√25-√24

=√25-√2=5-√2.Câu b tương tự

\(\frac{1}{\text{ }\sqrt{\frac{3}{5}}+\sqrt{\frac{3}{7}}+1}=\frac{1}{\frac{\sqrt{3.7}+\sqrt{3.5}+\sqrt{5.7}}{\sqrt{5.7}}}=\frac{\sqrt{35}}{\sqrt{21}+\sqrt{35}+\sqrt{15}}\)

Tương tự :

 \(\frac{1}{\sqrt{\frac{5}{3}}+\sqrt{\frac{5}{7}}+1}=\frac{\sqrt{21}}{\sqrt{35}+\sqrt{15}+\sqrt{21}}\)

 

\(\frac{1}{\sqrt{\frac{7}{3}}+\sqrt{\frac{7}{5}}+1}=\frac{\sqrt{15}}{\sqrt{21}+\sqrt{35}+\sqrt{15}}\)

Bây giờ chỉ việc cộng lại chung mẫu

Kq ; 1 

\(=\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{\sqrt{97}-\sqrt{99}}{97-99}\)

\(=\frac{\sqrt{3}-\sqrt{5}+\sqrt{5}-\sqrt{7}+...+\sqrt{97}-\sqrt{99}}{-2}\)

\(=\frac{\sqrt{3}-\sqrt{99}}{-2}=\frac{\sqrt{99}-\sqrt{3}}{2}\)

= \(\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{\sqrt{97}-\sqrt{99}}{97-99}\) = \(\frac{-1}{2}.\left(\sqrt{3}-\sqrt{5}+\sqrt{5}-\sqrt{7}+...+\sqrt{97}-\sqrt{99}\right)\)

= \(-\frac{1}{2}.\left(\sqrt{3}-\sqrt{99}\right)\) = \(\frac{3\sqrt{11}-\sqrt{3}}{2}\)

Xét biểu thức phụ : 1 (2n+3)√2n+1+(2n+1)√2n+3 = 1 √2n+1.√2n+3(√2n+1+√2n+3) = √2n+3−√2n+1 √2n+1.√2n+3[(2n+3)−(2n+1)] = √2n+3−√2n+1 2√2n+1.√2n+3 = 1 2 ( 1 √2n+1 − 1 √2n+3 )với n≥1 Áp dụng : S= 1 3√1+1√3 + 1 3√5+5√3 + 1 5√7+7√5 +...+ 1 101√103+103√101 = 1 2 ( 1 √1 − 1 √3 )+ 1 2 ( 1 √3 − 1 √5 )+ 1 2 ( 1 √5 − 1 √7 )+...+ 1 2 ( 1 √101 − 1 √103 ) = 1 2 (1− 1 √3 + 1 √3 − 1 √5 + 1 √5 − 1 √7 +...+ 1 √101 − 1 √103 ) = 1 2 (1− 1 √103 )

sai đề! P/S cuối phải là 2017