A=\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)............\left(\frac{1}{2017^2}-1\right)\left(\frac{1}{2018^2}-1\right)\)
B=\(-\frac{1}{2}\)
So sánh A và B
Cho A = \(\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2017^2}-1\right)\cdot\left(\frac{1}{2018^2}-1\right)\) và B = \(-\frac{1}{2}\)
Hãy so sánh A và B
cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2016^2}-1\right)\left(\frac{1}{2017^2}-1\right)\)và b=-1/2
Hãy so sánh A với B
Ta có:
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)
\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)
\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)
\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)
\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)
\(A=\frac{1.2018}{2017.2}\)
\(A=\frac{1009}{2017}\)
Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)
\(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)
Vậy A>B
Cho \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right).........\left(\frac{1}{2018^2}-1\right)\)
B= \(-\frac{1}{2}\)
So sánh A và B
â , tính M = \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right).......\left(1+\frac{1}{2017}\right)\left(1+\frac{1}{2018}\right)\)
b , Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2017}-\frac{1}{2018}\)
c , B = \(\frac{1}{1010}+\frac{1}{1011}+.....+\frac{1}{2017}+\frac{1}{2018}.tinh\left(\frac{A}{B}\right)^{2018}\)
a, \(M=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{3.4...2019}{2.3...2018}=\frac{2019}{2}\)
b, c cùng 1 câu phải k
ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)
\(\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2018}=1^{2018}=1\)
A,\(M=\frac{3}{2}\cdot\frac{4}{3}....\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{4\cdot3...2019}{2\cdot3...2018}=\frac{2019}{2}\)
NHA
HỌC TỐT
cho A=\(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)......\left(\frac{1}{2018^2}-1\right)\) và B= \(-\frac{1}{2}\).So sánh A và B
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)..........\left(\frac{1}{2018^2}-1\right)\)
Ta có :
\(\frac{1}{2^2}-1>-\frac{1}{2}\)
\(\frac{1}{3^2}-1>-\frac{1}{2}\)
...........
\(\frac{1}{2018^2}-1>\frac{1}{2}\)
\(\Rightarrow A>B\)
1. \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)
So sánh \(B\) với \(\frac{1}{4}\)
2. SO sánh \(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\) và \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
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 !!
Cho A= \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).......\left(\frac{1}{2018^2}-1\right)\)
B= \(-\frac{1}{2}\)
So sánh A và B
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2018^2}-1\right)\)
\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2018^2}\right)\)
\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot...\cdot\frac{4072323}{2018\cdot2018}\)
\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\cdot...\cdot\left(2017\cdot2019\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\cdot...\cdot\left(2018\cdot2018\right)}\)
\(-A=\frac{\left(1\cdot2\cdot...\cdot2017\right)\left(3\cdot4\cdot...\cdot2019\right)}{\left(2\cdot3\cdot...\cdot2018\right)\left(2\cdot3\cdot...\cdot2018\right)}\)
\(-A=\frac{1\cdot2019}{2018\cdot2}\)
\(-A=\frac{2019}{4036}\)
\(A=-\frac{2019}{4036}< -\frac{1}{2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2018^2}-1\right)\)
\(\Rightarrow A=\left(\frac{1}{2^2}-\frac{2^2}{2^2}\right)\left(\frac{1}{3^2}-\frac{3^2}{3^2}\right)...\left(\frac{1}{2018}-\frac{2018^2}{2018^2}\right)\)
\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}....\frac{-4072323}{2018^2}\)
\(\Rightarrow\frac{-\left(3.8.....4072323\right)}{\left(2.3.4...2018\right).\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-\left(1.3.2.4....2017.2019\right)}{\left(2.3.4...2018\right)\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-\left(\left(1.2.3...2017\right).\left(3.4.5..2019\right)\right)}{\left(2.3...2018\right)\left(2.3.4..2018\right)}\)
\(\Rightarrow A=\frac{-2019}{2018.2}< -\frac{2018}{2018.2}=\frac{-1}{2}\)
\(\Rightarrow A< \frac{-1}{2}\)
P/s: mk ko copy baì của bn uyên đâu nha
Cho A=\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2013^2}-1\right)..\left(\frac{1}{2014^2}-1\right)\&B=\frac{1}{2}\) so sánh A và B
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
Thực hiện phép tính:
a, A= \(\frac{1}{6}.\left(-2\frac{3}{5}\right)+1\frac{2}{3}.\left(\frac{-13}{5}\right)\)
b, B=\(\frac{5}{7}:\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{5}{7}:\left(\frac{1}{5}-\frac{1}{6}\right)\)
c, C= \(2^3+3.\left(\frac{2017}{2018}\right)^0.\left(\frac{1}{2}\right)^{-2}:4+[\left(-2\right)^2:\frac{1}{2}].\left(\frac{-1}{2}\right)^3\)
giúp vs, mik cần gấp
A= E387E4837
B = 883433
C = UỲUWFHQWURY48E3947