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Ta có A-1=2016/20^10-1
B-1= 2016/20^10-3
Suy ra a-1<B-1=>A<B
\(A=\frac{2015^{20}-6+6}{2015^{20}-6}=\frac{2015^{20}-6}{2015^{20}-6}+\frac{6}{2015^{20}-6}=1+\frac{6}{2015^{20}-6}\)
\(B=\frac{2015^{20}-2+6}{2015^{20}-2}=\frac{2015^{20}-2}{2015^{20}-2}+\frac{6}{2015^{20}-2}=1+\frac{6}{2015^{20}-2}\)
Vì \(2015^{20}-6\frac{6}{2015^{20}-2}\)
=> A > B
Vì A=201520-201519=201520-19=2015
B=201510-20159=201510-9=2015
Vậy A và B đều bằng nhau
Tích nha
c) E=2010-1+2/2010-1=1+2/210-1
F=2010-3+2/2010-3=1+2/2010-3
Vì 2/210-1>2/2010-3 nên E>F
a) Ta có: 10A=1020+10/1020+1
=1020+1/1020+1-9/1020+1
=1-9/1020+1
10B=1021+10/1021+1
=1021+1/1021+1-9/1021+1
=1-9/1021+1
Vì 9/1020+1>9/1021+1 nên A>B
(Còn nữa)
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>1;B<1
A>B
Ta có
\(A=\frac{20^{2015}+1}{20^{2015}-1}=\frac{20^{2015}-1+2}{20^{2015}-1}=1+\frac{2}{20^{2015}-1}\)
\(B=\frac{20^{2015}-1}{20^{2015}-3}=\frac{20^{2015}-3+2}{20^{2015}-3}=1+\frac{2}{20^{2015}-3}\)
Vì \(1+\frac{2}{20^{2015}-1}< 1+\frac{2}{20^{2015}-3}\)
\(\Rightarrow A< B\)