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:

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

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

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

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

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

\(=-6-3\sqrt{3}\)

1: ta có: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)

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

\(=2\sqrt{2}+\sqrt{5}+1\)

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

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

\(=4\sqrt{2}\)

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow A=1+\dfrac{1}{2.3}+1+\dfrac{1}{3.4}+....+1+\dfrac{1}{2021.2022}\)

\(=2020+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2021.2022}\)

\(=2020+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=2020+\dfrac{1}{2}-\dfrac{1}{2022}=...\)

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

Cmttt ta được:

\(A=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}+1+\dfrac{1}{2021}-\dfrac{1}{2022}\\ A=2020+\dfrac{1}{2}-\dfrac{1}{2022}=2020+\dfrac{505}{1011}=...\)

Ta chứng minh được công thức \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\)

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

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

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

\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)

\(=\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{4}+1+\dfrac{1}{2016}-\dfrac{1}{2017}+1+\dfrac{1}{2017}-\dfrac{1}{2018}\)

=>A là số hữu tỉ (ĐPCM)

4: Ta có: \(\dfrac{1}{3+\sqrt{5}}-\dfrac{1}{3-\sqrt{5}}\)

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

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

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

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

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