Tính giá trị của tổng
\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+........+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}.\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
21 tháng 7 2017
Ta có \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\Rightarrow A^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)
= \(\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}=\left[\frac{a^2+a+1}{a\left(a+1\right)}\right]^2\Rightarrow A=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a\left(a+1\right)}\)
= \(1+\frac{1}{a}-\frac{1}{a+1}\)
rồi bạn thay vào tổng trên là xong
5 tháng 7 2016
Xét : Với mọi \(x\in N^{\text{*}}\) , ta có : \(\frac{1}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}=\frac{1}{\sqrt{x\left(x+1\right)}\left(\sqrt{x}+\sqrt{x+1}\right)}=\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x\left(x+1\right)}}=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\)
Áp dụng vào tính : \(M=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
AH
Akai Haruma
Giáo viên
17 tháng 6 2019
Lời giải:
\(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}=\sqrt{1+2.\frac{1}{a}+\frac{1}{a^2}+\frac{1}{(a+1)^2}-\frac{2}{a}}\)
\(=\sqrt{(1+\frac{1}{a})^2+\frac{1}{(a+1)^2}-\frac{2}{a}}=\sqrt{\frac{(a+1)^2}{a^2}+\frac{1}{(a+1)^2}-2.\frac{a+1}{a}.\frac{1}{a+1}}\)
\(=\sqrt{(\frac{a+1}{a}-\frac{1}{a+1})^2}=|\frac{a+1}{a}-\frac{1}{a+1}|=|1+\frac{1}{a}-\frac{1}{a+1}|\)
b)
Áp dụng công thức trên vào bài toán:
\(B=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+....+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)
\(=|1+\frac{1}{1}-\frac{1}{2}|+|1+\frac{1}{2}-\frac{1}{3}|+....+|1+\frac{1}{99}-\frac{1}{100}|\)
\(=99+(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100})\)
\(=99+1-\frac{1}{100}=100-\frac{1}{100}\)
Y
17 tháng 6 2019
Sai đề nha bn \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\)
\(A=\sqrt{\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}\)\(=\sqrt{\frac{a^2\left(a+1\right)^2+2a^2+2a+1}{a^2\left(a+1\right)^2}}\)
\(=\sqrt{\frac{\left[a\left(a+1\right)^2\right]+2a\left(a+1\right)+1}{a^2\left(a+1\right)^2}}\) \(=\sqrt{\frac{\left[a\left(a+1\right)+1\right]^2}{a^2\left(a+1\right)^2}}\)
\(=\frac{a\left(a+1\right)+1}{a\left(a+1\right)}=1+\frac{1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)
Áp dụng kết quả trên ta có :
\(B=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{99}-\frac{1}{100}\)
\(=99+1-\frac{1}{100}=\frac{9999}{100}\)
6 tháng 8 2017
Với mọi n thuộc N ta có :
\(\sqrt{\frac{1}{1^2}+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+\frac{2}{n}-\frac{2}{n\left(n+1\right)}-\frac{2}{\left(n+1\right)}}\)
\(=\sqrt{\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2}=1+\frac{1}{n}-\frac{1}{n+1}\)
Áp dụng ta được :
\(S=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+....+\left(1+\frac{1}{99}-\frac{1}{100}\right)\)
\(=98+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=98+\frac{1}{2}-\frac{1}{100}=\frac{9849}{100}\)
WH
10 tháng 8 2020
fix lại tý nha !! dòng cuối :
Vì vậy : ... \(=99.1+\frac{1}{1}-\frac{1}{100}=\frac{9999}{100}\)
1 tháng 10 2018
Với a , b , c là số hữu tỉ t/m a = b + c ta luôn có \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right|\)
Thật vậy : \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2-2\left(\frac{1}{bc}-\frac{1}{ac}-\frac{1}{ab}\right)}\)
\(=\sqrt{\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2-\frac{2.abc\left(a-b-c\right)}{a^2b^2c^2}}\)(quy đồng lên )
\(=\sqrt{\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2}\left(\text{do a-b-c=0}\right)\)
\(=\left|\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right|\)
Áp dụng ta được \(S=\left|\frac{1}{2}-\frac{1}{1}-1\right|+\left|\frac{1}{3}-\frac{1}{2}-1\right|+...+\left|\frac{1}{100}-\frac{1}{99}-1\right|\)
\(=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+1+1+...+1\right)+\left(1+\frac{1}{2}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{100}\right)\)
(có 99 số 1)
\(=99+1-\frac{1}{100}\)
\(=100-\frac{1}{100}=\frac{9999}{100}\)
Bạn hãy chứng minh đẳng thức phụ sau : \(\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=\left|1-\frac{1}{k}+\frac{1}{k+1}\right|\)
Áp dụng : \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}=\left(1+1-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+...+\left(1+\frac{1}{99}-\frac{1}{100}\right)\)\(=1.99+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100-\frac{1}{100}\)
Với a \(\in\)N*, ta có:
\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=\sqrt{\frac{a^2.\left(a+1\right)^2}{a^2.\left(a+1\right)^2}+\frac{\left(a+1\right)^2}{a^2.\left(a+1\right)^2}+\frac{a^2}{a^2.\left(a+1\right)^2}}\)
\(=\sqrt{\frac{\left[a.\left(a+1\right)\right]^2+\left(a+1\right)^2+a^2}{\left[a.\left(a+1\right)\right]^2}}=\sqrt{\frac{\left[a.\left(a+1\right)\right]^2+a^2+2a+1+a^2}{\left[a.\left(a+1\right)\right]^2}}\)
\(=\sqrt{\frac{\left[a.\left(a+1\right)\right]^2+2a^2+2a+1}{\left[a.\left(a+1\right)\right]^2}}=\sqrt{\frac{\left[a.\left(a+1\right)\right]^2+2.\left(a^2+a\right)+1}{\left[a.\left(a+1\right)\right]^2}}\)
\(=\sqrt{\frac{\left[a.\left(a+1\right)\right]^2+2.a.\left(a+1\right).1+1^2}{\left[a.\left(a+1\right)\right]^2}}=\sqrt{\frac{\left[a.\left(a+1\right)+1\right]^2}{\left[a.\left(a+1\right)\right]^2}}\)
\(=\sqrt{\left[\frac{a.\left(a+1\right)+1}{a.\left(a+1\right)}\right]^2}=\frac{a.\left(a+1\right)+1}{a.\left(a+1\right)}=\frac{a.\left(a+1\right)}{a.\left(a+1\right)}+\frac{1}{a.\left(a+1\right)}\)
\(=1+\frac{a+1-a}{a.\left(a+1\right)}=1+\frac{a+1}{a.\left(a+1\right)}-\frac{a}{a.\left(a+1\right)}=a+\frac{1}{a}-\frac{1}{a+1}\)
=>\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=1+\frac{1}{a}-\frac{1}{a+1}\)
Thay a=1,2,...99
=>\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=1+1-\frac{1}{2}\)
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=1+\frac{1}{2}-\frac{1}{3}\)
............................................................
\(\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}=1+\frac{1}{99}-\frac{1}{100}\)
=>\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)
\(=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{99}-\frac{1}{100}\)
\(=1+1+...+1-\frac{1}{100}\)
\(=100-\frac{1}{100}\)
\(=\frac{9999}{100}\)
Vậy \(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}=\frac{9999}{100}\)