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AH
Akai Haruma
Giáo viên
31 tháng 5

Lời giải:
Xét số hạng tổng quát:
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=\sqrt{\frac{n^2+1}{n^2}+\frac{1}{(n+1)^2}}\\ =\sqrt{\frac{(n+1)^2}{n^2}-\frac{2n}{n^2}+\frac{1}{(n+1)^2}}\\ =\sqrt{\frac{(n+1)^2}{n^2}-\frac{2}{n}+\frac{1}{(n+1)^2}}\\ =\sqrt{(\frac{n+1}{n}-\frac{1}{n+1})^2}=\frac{n+1}{n}-\frac{1}{n+1}=1+\frac{1}{n}-\frac{1}{n+1}\)

Do đó:

\(C=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+....+1+\frac{1}{2018}-\frac{1}{2019}\\ =(1+1+...+1)+(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2018})-(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019})\\ =2018+1-\frac{1}{2019}=2019-\frac{1}{2019}\)

29 tháng 5

đáp án : 2019-1/2019

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

AH
Akai Haruma
Giáo viên
7 tháng 8 2021

1.

Đặt biểu thức là $A$

Ta thấy:

$\frac{1}{1+\sqrt{2}}=\frac{\sqrt{2}-1}{(1+\sqrt{2})(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$

Tương tự với các phân số còn lại và công theo vế thì:

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

$=\sqrt{2019}-1$

 

AH
Akai Haruma
Giáo viên
7 tháng 8 2021

2.

$\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt{5.3}+3}+\sqrt{3-2\sqrt{3.1}+1}$

$=\sqrt{(\sqrt{5}-\sqrt{3})^2}+\sqrt{(\sqrt{3}-1)^2}$

$=|\sqrt{5}-\sqrt{3}|+|\sqrt{3}-1|$

$=\sqrt{5}-\sqrt{3}+\sqrt{3}-1=\sqrt{5}-1$

AH
Akai Haruma
Giáo viên
17 tháng 7 2018

Lời giải:

Xét \(1+\frac{1}{n^2}+\frac{1}{(n+1)^2}=(1+\frac{1}{n})^2-\frac{2}{n}+\frac{1}{(n+1)^2}\)

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

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

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

\(\Rightarrow \sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=1+\frac{1}{n(n+1)}\)

Do đó:

\(A=1+\frac{1}{1.2}+1+\frac{1}{2.3}+...+1+\frac{1}{2018.2019}\)

\(=2018+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)

\(=2018+\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{2019-2018}{2018.2019}=2018+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(=2019-\frac{1}{2019}\)

29 tháng 11 2018

co cong thuc \(\sqrt{\dfrac{1}{1}+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}=\dfrac{1}{1}+\dfrac{1}{a}-\dfrac{1}{a+1}\) ban tu chung minh nha

4 tháng 10 2018

Bài 1:Với mọi n∈N*,ta có:

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

Do đó :

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

Bài 2: 

\(A=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)

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

=10

28 tháng 10 2022

\(=1+\dfrac{1}{1}-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2018}-\dfrac{1}{2019}\)

=2017-1/2019

NV
22 tháng 1 2019

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

\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)

\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)

\(\Rightarrow A=\dfrac{3n}{6\left(3n+2\right)}=\dfrac{n}{6n+4}\)

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

\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{3.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)

\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)

\(\Rightarrow B=\dfrac{n\left(n+2\right)}{3\left(2n+1\right)\left(2n+3\right)}\)

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

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

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

\(\Rightarrow C=1+\dfrac{1}{1}-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2018}-\dfrac{1}{2019}\)

\(\Rightarrow C=2019-\dfrac{1}{2019}\)

22 tháng 1 2019

@Luân Đào @Nguyễn Việt Lâm

28 tháng 7 2018

\(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{a}\right)^2-\dfrac{2}{a}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(\dfrac{a+1}{a}\right)^2-2.\dfrac{a+1}{a}.\dfrac{a}{a+1}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{a}-\dfrac{1}{a+1}\right)^2}=1+\dfrac{1}{a}-\dfrac{1}{a+1}\left(a>0\right)\)

Áp dụng điều này vào bài toán , ta có :

\(P=\sqrt{1^2+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1^2+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1^2+\dfrac{1}{2018^2}+\dfrac{1}{2019^2}}=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2018}-\dfrac{1}{2019}=2017+\dfrac{1}{2}-\dfrac{1}{2019}\)

AH
Akai Haruma
Giáo viên
10 tháng 8 2018

Lời giải:

Xét số hạng tổng quát:

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

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

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

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

Do đó:

\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)

\(=1-\frac{1}{\sqrt{2019}}\)

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)

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

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

\(=4\sqrt{2}\)