Tính
A =\(\frac{1}{2}\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2015.2017}\right)\)
B = 2x2-3x+5 với /x/ = \(\frac{1}{2}\)
C=15(y2x - x2y) với x-y=0
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
4 tháng 11 2016
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2015.2017}\right)\)
\(=\frac{1}{2}.\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}...\frac{2015.2017+1}{2015.2017}\)
\(=\frac{1}{2}.\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2016.2016}{2015.2017}\)
\(=\frac{1}{2}.\frac{2.3.4...2016}{1.2.3...2015}.\frac{2.3.4...2016}{3.4.5...2017}\)
\(=\frac{1}{2}.2016.\frac{2}{2017}=\frac{2016}{2017}\)
4 tháng 11 2016
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2015.2017}\right)\)
\(=\frac{1}{2}.\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}...\frac{2015.2017+1}{2015.2017}\)
\(=\frac{1}{2}.\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2016.2016}{2015.2017}\)
\(=\frac{1}{2}.\frac{2.3.4...2016}{1.2.3...2015}.\frac{2.3.4...2016}{3.4.5...2017}\)
\(=\frac{1}{2}.2016.\frac{2}{2017}=\frac{2016}{2017}\)
9 tháng 4 2017
2A=\(\left(1+\frac{1}{3}\right)\)\(\left(1+\frac{1}{8}\right)\)\(\left(1+\frac{1}{15}\right)\)\(.......\)\(\left(1+\frac{1}{4064255}\right)\)
2A = \(\frac{4}{3}\)\(.\)\(\frac{9}{8}\)\(.\)\(\frac{16}{15}\)\(......\)\(\frac{4064256}{4064255}\)
2A = \(\frac{2.2}{1.3}\)\(.\)\(\frac{3.3}{2.4}\)\(.\)\(\frac{4.4}{3.5}\)\(......\)\(\frac{2016.2016}{2015.2017}\)
2A = \(\frac{2.3.4....2016}{1.2.3.....2015}\)\(.\)\(\frac{2.3.4....2016}{3.4.5....2017}\)
2A = \(\frac{2016}{1}\)\(.\)\(\frac{2}{2017}\)
2A = \(\frac{4032}{2017}\)
A = \(\frac{4032}{2017}\)\(:2\)
A = \(\frac{2016}{2017}\)
6 tháng 2 2022
a: \(\Leftrightarrow\dfrac{x-214}{86}-1+\dfrac{x-132}{84}-2+\dfrac{x-54}{82}-3=0\)
=>x-300=0
hay x=300
PN
11 tháng 11 2015
Công thức tống quát:
\(1+\frac{1}{\left(n-1\right)\left(n+1\right)}=1+\frac{1}{n^2-1}=\frac{n^2-1+1}{n^2-1}=\frac{n^2}{n^2-1}\)
Theo đó, ta có:
\(1+\frac{1}{1.3}=1+\frac{1}{\left(2-1\right)\left(2+1\right)}=\frac{2^2}{2^2-1}\)
\(1+\frac{1}{2.4}=1+\frac{1}{\left(3-1\right)\left(3+1\right)}=\frac{3^2}{3^2-1}\)
\(1+\frac{1}{3.5}=\frac{1}{\left(4-1\right)\left(4+1\right)}=\frac{4^2}{4^2-1}\)
\(....................\)
\(1+\frac{1}{2015.2017}=1+\frac{1}{\left(2016-1\right)\left(2016+1\right)}=\frac{2016^2}{2016^2-1}\)
Nhân lần lượt các đẳng thức trên, ta được:
\(S=\frac{\left(2.3.4....2016\right)^2}{\left(2^2-1\right)\left(3^2-1\right)\left(4^2-1\right)...\left(2016^2-1\right)}=\frac{2^2.3^2.4^2...2016^2}{\left(1.3\right)\left(2.4\right)\left(3.5\right)....\left(2015.2017\right)}=\frac{2^2.3^2.4^2...2016^2}{1.2.3^2.4^2.5^2...2014^2.2015^2.2016.2017}=\frac{2.2016}{2017}\)
1 tháng 11 2019
\(B=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)...\left(\frac{n.\left(n+2\right)+1}{n.\left(n+2\right)}\right)\)
\(=\left(\frac{2^2}{1.3}\right).\left(\frac{3^2}{2.4}\right).\left(\frac{4^2}{3.5}\right)...\left(\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\right)\)
\(=\frac{2.3.4...\left(n+1\right)}{1.2.3...n}.\frac{2.3.4...\left(n+1\right)}{3.4.5...\left(n+2\right)}\)
\(=\frac{\left(n+1\right)}{1}.\frac{2}{\left(n+2\right)}\)
\(=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}=2.\frac{n+1}{n+2}< 2\)(vì \(\frac{n+1}{n+2}< 1\))
Vậy B < 2
19 tháng 9 2019
Ta có:
\(1+\frac{1}{1.3}=\frac{4}{1.3}=\frac{2^2}{1.3}\)
\(1+\frac{1}{2.4}=\frac{9}{2.4}=\frac{3^2}{2.4}\)
\(1+\frac{1}{3.5}=\frac{16}{3.5}=\frac{4^2}{3.5}\)
...
\(1+\frac{1}{n\left(n+2\right)}=\frac{n^2+2n+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
=>
\(B=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2^2.3^2.4^2...\left(n+1\right)^2}{1.2.3^2.4^2...\left(n+1\right)\left(n+2\right)}=\frac{2.\left(n+1\right)}{1.\left(n+2\right)}\)
\(=\frac{2\left(n+2\right)-2}{n+2}=2-\frac{2}{n+2}< 2\)
Vậy B < 2
a,A=\(\frac{1}{2}.\left(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2016.2016}{2015.2017}\right)=\frac{1}{2}.\left(\frac{2.3.4...2016}{1.2....2015}.\frac{2.3.4...2016}{3.4....2017}\right)=\frac{1}{2}.\left(\frac{2016.2}{2017}\right)=\frac{4032}{4034}=\frac{2016}{2017}\)
Hok tốt
\(\left|x\right|=\frac{1}{2}\Rightarrow x=\orbr{\begin{cases}\frac{1}{2}\\-\frac{1}{2}\end{cases}}\)
TH1:\(x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{3}{2}+5=4\)
TH2:\(x=\frac{-1}{2}\)
\(\Rightarrow\frac{1}{2}+\frac{3}{2}+5=7\)
Vậy