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\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)..........\left(\frac{1}{2018^2}-1\right)\)
Ta có :
\(\frac{1}{2^2}-1>-\frac{1}{2}\)
\(\frac{1}{3^2}-1>-\frac{1}{2}\)
...........
\(\frac{1}{2018^2}-1>\frac{1}{2}\)
\(\Rightarrow A>B\)
Bài 1:
ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)
\(B=\frac{1}{2^2}-\frac{1}{100^2}\)
\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)
\(\Rightarrow B< \frac{1}{4}\)
Bài 2:
ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
Học tốt nhé bn !!
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2018^2}-1\right)\)
\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2018^2}\right)\)
\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot...\cdot\frac{4072323}{2018\cdot2018}\)
\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\cdot...\cdot\left(2017\cdot2019\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\cdot...\cdot\left(2018\cdot2018\right)}\)
\(-A=\frac{\left(1\cdot2\cdot...\cdot2017\right)\left(3\cdot4\cdot...\cdot2019\right)}{\left(2\cdot3\cdot...\cdot2018\right)\left(2\cdot3\cdot...\cdot2018\right)}\)
\(-A=\frac{1\cdot2019}{2018\cdot2}\)
\(-A=\frac{2019}{4036}\)
\(A=-\frac{2019}{4036}< -\frac{1}{2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2018^2}-1\right)\)
\(\Rightarrow A=\left(\frac{1}{2^2}-\frac{2^2}{2^2}\right)\left(\frac{1}{3^2}-\frac{3^2}{3^2}\right)...\left(\frac{1}{2018}-\frac{2018^2}{2018^2}\right)\)
\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}....\frac{-4072323}{2018^2}\)
\(\Rightarrow\frac{-\left(3.8.....4072323\right)}{\left(2.3.4...2018\right).\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-\left(1.3.2.4....2017.2019\right)}{\left(2.3.4...2018\right)\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-\left(\left(1.2.3...2017\right).\left(3.4.5..2019\right)\right)}{\left(2.3...2018\right)\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-2019}{2018.2}< -\frac{2018}{2018.2}=\frac{-1}{2}\)
\(\Rightarrow A< \frac{-1}{2}\)
P/s: mk ko copy baì của bn uyên đâu nha
Ta có
\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right).....\left(1^2-2014^2\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)
\(\Leftrightarrow A=\frac{\left(-1\right)3\left(-2\right)4.....\left(-2013\right)2015}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)
\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right)...\left(-2013\right)\right]\left(3.4.5...2015\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)
\(\Leftrightarrow A=\frac{\left(-1\right)2015}{2014.2}=-\frac{2015}{4028}< -\frac{2014}{4028}=-\frac{1}{2}\)
=> A<-1/2
A có: \(\frac{2014-2}{3-2}+1=2013\) ( thừa số )
Ta thấy mỗi thừa số của A đều có dạng \(\frac{1}{n^2}-1\)với \(n\inℕ^∗\)và \(n>1\)
Có \(\frac{1}{n^2}< 1\Rightarrow\frac{1}{n^2}-1< 1-1=0\)
=> Mỗi thừa số của A đều nhỏ hơn 0
=> A là tích của 2013 thừa số nhỏ hơn 0
Mà 2013 là số lẻ
=> A < 0
Mà B = \(\frac{1}{2}\)> 0
=> A < B
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2017^2}-1\right)\left(\frac{1}{2018^2}-1\right)\)
\(A=\frac{\left(1-2^2\right)\left(1-3^2\right)\left(1-4^2\right)...\left(1-2018^2\right)}{2^23^24^2...2018^2}\)
\(A=\frac{-1\cdot3\cdot\left(-2\right)\cdot4\cdot\left(-3\right)\cdot5\cdot...\cdot\left(-2016\right)\cdot2018}{2018!^2}\)
\(A=\frac{2016!\cdot3\cdot4\cdot5\cdot...\cdot2018}{2018!^2}=\frac{2016!\cdot2018!}{2018!^2\cdot2!}=\frac{2016!}{2!2018!}=\frac{1}{2!\cdot2017\cdot2018}>0>-\frac{1}{2}=B\)
A = (1/2+1)(1/2-1)(1/3+1)(1/3-1)....(1/2018+1)(1/2018-1) đặt các tích phần tử có dấu + là X, tích các phần tử có dấu - là Y => A= X.Y
X = 3/2.4/3.5/4.....2019/2018 = 2019/2
Y= (-1/2)(-2/3)(-3/4)...(-2017/2018) = -1/2018 (tích của 2017 số <0)
A= X.Y = -2019/2018.1/2 < B= -1/2