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2 tháng 9 2018

Chứng minh công thức:

\(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}\)

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

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

\(=\sqrt{\dfrac{a^4+2a^3+a^2+a^2+2a+1+a^2}{a^2\left(a+1\right)^2}}\)

\(=\sqrt{\dfrac{a^4+2a^3+3a^2+2a+1}{a^2\left(a+1\right)^2}}\)

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

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

\(=\dfrac{a^2+a+1}{a\left(a+1\right)}\)

\(=\dfrac{a\left(a+1\right)+1}{a\left(a+1\right)}\)

\(=1+\dfrac{1}{a\left(a+1\right)}\)

\(=1+\dfrac{1}{a}-\dfrac{1}{a+1}\)

Áp dụng công thức ta có:

\(C=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+...+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{1999^2}+\dfrac{1}{2000^2}}\)

\(=1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{1999}-\dfrac{1}{2000}\)

\(=2000-\dfrac{1}{2000}=\dfrac{1999}{2000}\)

3 tháng 9 2018

hình như \(\dfrac{1}{2}\) không bị trừ bn ạ

7 tháng 8 2017

Ta chứng minh công thức:

\(1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}=\left(1+\dfrac{1}{n}+\dfrac{1}{n+1}\right)^2\) bằng cách quy đồng biểu thức ở vế phải rồi áp dụng vào bài tập

10 tháng 5 2017

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

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

10 tháng 7 2017

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

5 tháng 7 2017

\(C=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2.2+\sqrt{5+2\sqrt{5}+1}}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2.2-\sqrt{5-2\sqrt{5}+1}}=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{4+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{4-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{2.}\left(\dfrac{3+\sqrt{5}}{5+\sqrt{5}}+\dfrac{3-\sqrt{5}}{5-\sqrt{5}}\right)=\sqrt{2}\left(\dfrac{\left(3+\sqrt{5}\right)\left(5-\sqrt{5}\right)+\left(3-\sqrt{5}\right)\left(5+\sqrt{5}\right)}{\left(5+\sqrt{5}\right)\left(5-\sqrt{5}\right)}\right)=\sqrt{2}.\dfrac{15+5\sqrt{5}-3\sqrt{5}-5+15-5\sqrt{5}+3\sqrt{5}-5}{25-5}=\sqrt{2}.\dfrac{20}{20}=\sqrt{2}\)

5 tháng 7 2017

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

\(=\dfrac{2+\sqrt{6}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2-\sqrt{6}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

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

\(=\dfrac{12-6\sqrt{2}}{9-3}=2-\sqrt{2}\)

b: Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)

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

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

23 tháng 6 2017

(bài 1) a) \(\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}\) = \(\dfrac{5-2\sqrt{6}-5-2\sqrt{6}}{25-24}\)

= \(\dfrac{-4\sqrt{6}}{1}\) = \(-4\sqrt{6}\)

b) \(\sqrt{6+2\sqrt{5}}-\dfrac{\sqrt{15}-\sqrt{3}}{\sqrt{3}}\) = \(\sqrt{\left(\sqrt{5}+1\right)^2}-\dfrac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{3}}\)

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

c) \(\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\) = \(\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{16}}\)

= \(\sqrt{6}.\sqrt{16}\) = \(4\sqrt{6}\)

d) \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)

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

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

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

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

e) \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}\) = \(\dfrac{4}{1+\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}\)

= \(\dfrac{4}{1+\sqrt{3}}-\sqrt{3}\) = \(\dfrac{4-\sqrt{3}-3}{1+\sqrt{3}}\) = \(\dfrac{1-\sqrt{3}}{1+\sqrt{3}}\)

= \(\dfrac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1-3}\) = \(\dfrac{1-2\sqrt{3}+3}{-2}\) = \(\dfrac{4-2\sqrt{3}}{-2}\)

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

23 tháng 6 2017

bài 2)

a)\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(a+b-2\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)

= \(\dfrac{a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}-2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)

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

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

b) \(\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)

= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

= \(\dfrac{2a-2}{4\sqrt{a}}.\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

= \(\dfrac{2\left(a-1\right)}{4\sqrt{a}}.\dfrac{-4a}{a-1}\) = \(-2\)

11 tháng 6 2021

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)

 

11 tháng 6 2021

cả 2 ý bạn trục căn thức ở mấu là xong nhé:

vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy

31 tháng 7 2018

a/ Ta có:

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

\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)

31 tháng 7 2018

a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)