Xét biểu thức phụ : \(\frac{1}{\left(2n+3\right)\sqrt{2n+1}+\left(2n+1\right)\sqrt{2n+3}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)

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

\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{2\sqrt{2n+1}.\sqrt{2n+3}}=\frac{1}{2}\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với \(n\ge1\)

Áp dụng : \(S=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{101\sqrt{103}+103\sqrt{101}}\)

\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}\right)+\frac{1}{2}\left(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}\right)+...+\frac{1}{2}\left(\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{101}}-\frac{1}{\sqrt{103}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{103}}\right)\)

DM CHƯA HỌC ĐẾN

Tất cả bằng 1 tin đi

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 )

\(Q=\left(\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{-\left(\sqrt{3}-1\right)}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)

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

\(S=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\sqrt{\frac{6+2\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}=\frac{\sqrt{10}-\sqrt{2}}{4}-\sqrt{\frac{\left(\sqrt{5}+1\right)^2}{4}}\)

\(=\frac{\sqrt{10}-\sqrt{2}}{4}-\frac{\sqrt{5}+1}{2}=\) bạn coi lại đề

\(T=\frac{4\left(1+\sqrt{3}\right)}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\frac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=\frac{4\left(1+\sqrt{3}\right)}{-2}-\sqrt{3}=-2-3\sqrt{3}\)

Bạn ơi cái này mk chỉ ghi cách làm và ct thôi nha 

đây dùng hàng đẳng thức (a-b)(a+b)=a^2-b^2

còn kia là công thức toán lớp 6

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

Tương tự:

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

.....

\(\frac{1}{\sqrt{2019}+\sqrt{2017}}=\frac{1}{2}\left(\sqrt{2019}-\sqrt{2017}\right)\)

Cộng các vế với nhau ta được:

\(S=\frac{1}{2}\left(\sqrt{2019}-\sqrt{1}\right)=\frac{1}{2}\left(\sqrt{2019}-1\right)\)

Bài này em dùng trục căn thức:

VD: \(\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}-1}{\left(\sqrt{3}\right)^2-1}=\frac{\sqrt{3}-1}{2}\)

Tương tự thì ta có:

\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)

\(=\frac{\sqrt{2019^2}-1}{2}=\frac{2019-1}{2}=1009\)

\(S=\frac{\sqrt{3}-1}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2019^2-\left(2019^2-2\right)}\)

\(S=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+...+\frac{\sqrt{2019^2}-\sqrt{2019^2-2}}{2}\)

\(S=\frac{1}{2}\left(\sqrt{3}-1+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+...+\sqrt{2019^2}-\sqrt{2019^2-2}\right)\)

\(S=\frac{1}{2}\left(-1+\sqrt{2019^2}\right)\)

\(S=\frac{\left(2019-1\right)}{2}=1009\)

\(S=\frac{1-\sqrt{3}}{1-3}+\frac{\sqrt{3}-\sqrt{5}}{3-5}+\frac{\sqrt{5}-\sqrt{7}}{5-7}+...+\frac{2019-\sqrt{2019^2-2}}{2019^2-2019^2-2}.\)

\(S=\frac{1-\sqrt{3}}{-2}+\frac{\sqrt{3}-\sqrt{5}}{-2}+\frac{\sqrt{5}-\sqrt{7}}{-2}+...+\frac{2019-\sqrt{2019^2-2}}{-2}.\)

\(-2S=1-\sqrt{3}+\sqrt{3}-\sqrt{5}+\sqrt{5}...+2019-\sqrt{2019^2-2}\)

\(-2S=1-\sqrt{2019^2-2}\Rightarrow S=\frac{\sqrt{2019^2-2}-1}{2}\)

cảm ơn bạn rất nhiều

\(1+3+5+...+\left(2n+1\right)=\left(n+1\right)^2\)

\(\Rightarrow S=\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+\left(n+1\right)}\)

\(=\frac{1}{1}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{\left(n+1\right)\left(n+2\right)}\)

\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n+1}-\frac{1}{n+2}\right)\)

\(=1+2\left(\frac{1}{2}-\frac{1}{n+2}\right)=\frac{2n+2}{n+2}\)

Viết ngược lại đề bài nha  rồi trục căn thức

\(2S=\sqrt{2013}-\sqrt{2011}+\sqrt{2011}-\sqrt{2009}+....+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}\)

\(S=\frac{\sqrt{2013}-\sqrt{3}}{2}\)