SO SANH:
\(A=\frac{10^9+1}{10^{10}+1}\)
\(B=\frac{10^7+1}{10^8+1}\)
Những câu hỏi liên quan
so sanh A,B .A=\(\frac{10^9+1}{10^8+1}\) ,,B\(\frac{10^{20}+1}{10^{21}+1}\)
Ta có 108 > 1
và 108 < 109
=) A > 1
Ta có : 1021 > 1020
1021 > 1
=) B < 1
Vậy A > B
So sanh A va B, biet :
a)\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
b)\(A=\frac{7^{10}}{1+7+7^2+...+7^9};B=\frac{5^{10}}{1+5+5^2+...+5^9}\)
\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
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cho A=10^7+1/10^8+1
B=10^8+1/10^9+1
so sanh A va B
các bạn giúp mình với nhanh lên nhé.Mình sẽ cho
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Bài 1:Tìm x biết Bài 2:So sánh
a, x+frac{1}{2}frac{3}{8}.frac{4}{5} a, Afrac{10^{10}-1}{10^{11}-1}vaBfrac{10^9-1}{10^{10}-1}
b, frac{5}{16}:x-frac{1}{4}frac{5}{8} b, B frac{10^{10}}{10^{10}+1}vaBfrac{10^{10}+1}{10^{10}+2}
c, frac{-1}{4}.x+frac{3}{7}.x2
d, frac{22}{9}-left(x+frac{1}{2}right)^2frac{7}{3}
e, left|frac{1}{4}-xright|+5frac{1}{8}6frac{1}{8}
Đọc tiếp
Bài 1:Tìm x biết Bài 2:So sánh
a, \(x+\frac{1}{2}=\frac{3}{8}.\frac{4}{5}\) a, \(A=\frac{10^{10}-1}{10^{11}-1}vaB=\frac{10^9-1}{10^{10}-1}\)
b, \(\frac{5}{16}:x-\frac{1}{4}=\frac{5}{8}\) b, B =\(\frac{10^{10}}{10^{10}+1}vaB=\frac{10^{10}+1}{10^{10}+2}\)
c, \(\frac{-1}{4}.x+\frac{3}{7}.x=2\)
d, \(\frac{22}{9}-\left(x+\frac{1}{2}\right)^2=\frac{7}{3}\)
e, \(\left|\frac{1}{4}-x\right|+5\frac{1}{8}=6\frac{1}{8}\)
So sánh:
a, A= \(\frac{10^8+2}{10^8-1}\) ; B= \(\frac{10^8}{10^8-3}\)
b, A= \(\frac{8^{10}+1}{8^{10}-1}\) ; B=\(\frac{8^{10}-1}{8^{10}-3}\)
c, A= \(\frac{100^9+4}{100^9-1}\): B= \(\frac{100^9+1}{100^9-4}\)
mk giải cho câu A rồi tự suy mấy câu khác nhé!
ta có : A = 10^8 + 2/10^8 - 1
=> A = 10^8 - 1 + 3/10^8 - 1
=> A = 1+ 3/10^8 - 1
B = 10^8/10^8 - 3
=> B = 10^8 - 3 + 3/10^8 - 3
=> B = 1+ 3/10^8 - 3
vì 3/10^8 - 1 < 3/10^8 - 3
=> 1 + 3/10^8 - 1 < 1 + 3/10^8 - 3
=> A < B
vậy A < B
cách này cô dạy mk đó
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bài 1 So sánha)Afrac{3}{8^3}+frac{7}{8^4} ; Bfrac{7}{8^3}+frac{3}{8^4}b)Afrac{10^{1992}+1}{10^{1991}+1};Bfrac{10^{1993}+1}{10^{1992}+1}c)Afrac{10^7+5}{10^4-8};Bfrac{10^8+6}{10^8-7}d)Afrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};Bfrac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}e)Afrac{2011}{2012}+frac{2012}{2013};Bfrac{2011+2012}{2012+2013}
Đọc tiếp
bài 1 So sánh
a)\(A=\frac{3}{8^3}+\frac{7}{8^4}\) ; \(B=\frac{7}{8^3}+\frac{3}{8^4}\)
b)\(A=\frac{10^{1992}+1}{10^{1991}+1};B=\frac{10^{1993}+1}{10^{1992}+1}\)
c)\(A=\frac{10^7+5}{10^4-8};B=\frac{10^8+6}{10^8-7}\)
d)\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
e)\(A=\frac{2011}{2012}+\frac{2012}{2013};B=\frac{2011+2012}{2012+2013}\)
so sanh A B biet A=\(\frac{10^{2018}+1}{10^{2017}+1}\);B=\(\frac{10^{2007}+1}{10^{2008+1}}\)
Ta có: \(10^{2018}>10^{2017}\Rightarrow10^{2018}+1>10^{2017}+1\Rightarrow A=\frac{10^{2018}+1}{10^{2017}+1}>1\) (1)
\(10^{2007}< 10^{2008}\Rightarrow10^{2007}+1< 10^{2008}+1\Rightarrow B=\frac{10^{2007}+1}{10^{2008}+1}< 1\) (2)
Từ (1) và (2) => A > B
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so sanh 2 PS sau
\(\frac{7^{10}+1}{7^{10}-1}\)va \(\frac{7^{10}-1}{7^{10}-3}\)
Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
B = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
a) So sánh A và B
b) Chứng minh A = \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
