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Bài 1 : dễ bạn tự làm được :)
Bài 2 :
Ta có :
\(B=\frac{2015+2016+2017}{2016+2017+2018}=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì :
\(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)
\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)
Nên \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015+2016+2017}{2016+2017+2018}\)
\(\Leftrightarrow\)\(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
Ta có : B = 2016 + 2017 + 2018 2015 + 2016 + 2017 = 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 Vì : 2016 2015 > 2016 + 2017 + 2018 2015 2017 2016 > 2016 + 2017 + 2018 2016 2018 2017 > 2016 + 2017 + 2018 2017 Nên 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 + 2018 2016 + 2016 + 2017 + 2018 2017 ⇔ 2016 2015 + 2017 2016 + 2018 2017 > 2016 + 2017 + 2018 2015 + 2016 + 2017 ⇔A > B Vậy A > B Chúc bạn học tốt ~
ta có:
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2017}}\)
\(\Rightarrow2A-A=2-\frac{1}{2^{2018}}\)
\(\Rightarrow A=\frac{2^{2019}-1}{2^{2018}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2018}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{2017}}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+........+\frac{1}{2^{2017}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{2018}}\right)\)
\(\Rightarrow A=2-\frac{1}{2^{2018}}\)
\(\Rightarrow A=\frac{2^{2019}-1}{2^{2018}}\)
\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)
\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)
2.
\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)
\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)
3.
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
4.
\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)
\(=\frac{1}{2}\)
mình chỉ làm được câu 3 thôi
có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)
\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)
\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)
\(=\frac{-1}{100}\)
P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
1) Đặt dãy trên là \(A\)
Theo bài ra ta có :
\(A=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)
\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)
2) \(A=\frac{5^{2018}-2017+1}{5^{2018}-2017}=\frac{5^{2018}-2017}{5^{2018}-2017}+\frac{1}{5^{2018}-2017}=1+\frac{1}{5^{2018}-2017}\)( 1 )
\(B=\frac{5^{2018}-2019+1}{5^{2018}-2019}=\frac{5^{2018}-2019}{5^{2018}-2019}+\frac{1}{5^{2018}-2019}=1+\frac{1}{5^{2018}-2019}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\)\(A=1+\frac{1}{5^{2018}-2017}< 1+\frac{1}{5^{2018}-2019}=B\)
\(\Rightarrow A< B\)
Vậy \(A< B.\)
1) Ta có B =
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) < \(\frac{1}{1.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)= \(\frac{99}{100}\)
=> B < 1 ( chứ không phải \(\frac{1}{2}\) bạn nhé)
Sai thì thôi chứ mk chỉ làm rờ thôi
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{2018}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2018}{2017}.\frac{2019}{2018}\)
\(=\frac{3.4.5.6.....2018.2019}{2.3.4.5....2017.2018}=\frac{2019}{2}\)
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{2018}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2019}{2018}\)
\(=\frac{3.4.5....2019}{2.3.4....2018}\)
\(=\frac{2019}{2}\)
Đặt \(B=1+2^2+2^3+...+2^{2016}\)
Gấp B lên 2 lần ta có :
2B = 2 . (1 + 2 + 22 + 23 + ... + 22016 )
2B = 2 + 22 + 23 + ... + 22017
2B - B = ( 2 + 22 + ... + 22017 ) - ( 1 + 2 + 22 + ... + 22016 )
B = 22017 - 2
Thay B vào ta có :
\(A=\frac{2^{2017}-2}{2^{2018}-2}\)
# Hoq chắc _ Baccanngon