so sanh A va B voi 0
A = 1. ( - 2 ) . 3 . ( - 4 ) ..... . 99 . ( -100 )
B = 1 . ( - 2 ) . 3 . ( - 4 ) . ... . (-98) . 99
so sanh A va B voi 0
A = 1 . ( - 2 ) . 3 .( - 4 ) . .... . 99 . ( - 100 )
B = 1 . ( - 2 ) . 3. ( - 4 ) . .... . ( - 98 ) . 99
So sanh A va B
A=(1+3+5+.......+99) / 50
B=( 2+4+6+.....+98)/ 49
A=1/1*2+1/2*3+1/3*4+......+1/99*100 so sanh voi 1
A = 1/1×2 + 1/2×3 + 1/3×4 + .. + 1/99×100
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
A = 1 - 1/100 < 1
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1-\frac{1}{100}< 1\)
=> ĐPCM
Ta có:
A = 1/1 x 2 + 1/2 x 3 + 1/3 x 4 + ..... + 1/99 x 100
A = 1- 1/2 + 1/2 - 1 /3 + 1/3 - 1/4 + ..... + 1/99 - 1/100
A = 1 - 1/100 < 1
nha bn
chúc bn học giỏi
so sanh A va B cho A=1+1+4^2+...+4^99;B=4^100
A = 1 + 4 + 4^2 + ... + 4^99
4A = 4 + 4^2 + 4^3 +... + 4^100
4A - A = 3A = ( 4 + 4^2 + 4^100 ) - ( 1 + 4 + 4^2 + 4^99 )
3A = 4^100 - 1
Ta thấy: 3A < B => A < B/3 ( đpcm )
k đúng nhé
A = 1 + 4 + 42 + ... + 499
4A = 4 + 42 + 43 + ... + 4100
4A - A = ( 4 + 42 + 43 + ... + 4100 ) - ( 1 + 4 + 42 + ... + 499 )
3A = 4100 - 1
A = \(\frac{4^{100}-1}{3}\)
Mà B = 4100
\(\Rightarrow\)A < B
cho A =1/2*3/4*5/6*...*99/100
B=2/3*4/5*6/7*...*100/101
C=1/2*2/3*4/5*...*98/99
a) so sanh A, B, C
b) Chung minh: A*C< A^2< 1/10
c) Chung minh: 1/15< A< 1/10
Lam giup minh di ai lam duoc minh tich dung cho
so sanh
a)\(A=\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+.....+\frac{5}{4^{99}}vaB=\frac{5}{3}\)
b)\(B=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.....+\frac{3^{98}+1}{3^{98}}vaA=100\)
Cho A = 100+1/99+2/98+...99/1
B = 100-1/2-2/3-3/4-...-98/99
Tính A/B
giai giup e voi
a,-1+3-5+7+...+97-99
b,1+2-3-4+...+97+98-99-100
a) (-1+3) + (-5+7) +(-9+11) + ...+ (-97+99)=2.50=100
b) 1+2-3-4+5+...+97+98-99-100
=1+(2-3-4+5)+...+(98-99-100+101)-101
=1+0+0+...+0-101
=-100
Tính A:B
a)A=98+1/2+1/3+1/4+...+1/99
B=2/3+4/3+5/4+...+100/99
b)A=2018+1/2+1/3+1/4+...+1/2019
B=3/2+4/3+3/4+...+2020/2019
c)A=99/1+98/2+97/3+...+2/98+1/99
B=1/2+1/3+1/4+...+1/100
Giải đầy đủ
a) \(A=98+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}=1\)
b) \(A=2018+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\)(có 2018 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{2019}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}=1\)
c) \(A=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)
\(A=99+\frac{98}{2}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào từng phân số)
\(A=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Và \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}=100\)
a)\(B=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{100}{99}\)
\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{99}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\right)\)
\(\Rightarrow B=98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
b)\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2019}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)\)
\(\Rightarrow B=2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}=1.\)
Vậy \(A:B=1.\)
c)\(A=\left(1+1+...+1\right)+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)
\(A=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+...+\left(1+\frac{2}{98}\right)+\left(1+\frac{1}{99}\right)\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)