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So sánh A = (9/11 - 0,81)^2005 và B = 1/(10)^4010
ta được A =B =0
chúc bạn học tốt
ơi bạn hoang thi kim hãy giải thích kặn kẻ hơn được không, nếu mình thấy đúng sẽ cho một k
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Ta có:\(\left(\frac{9}{11}-0,81\right)^{2005}\)=\(\left(\frac{9}{11}-\frac{81}{100}\right)^{2005}=\left(\frac{9}{1100}\right)^{2005}< \left(\frac{10}{1100}\right)^{2005}=\left(\frac{1}{110}\right)^{2005}\)
Mà \(\left(\frac{1}{110}\right)^{2005}< \left(\frac{1}{100}\right)^{2005}=\left[\left(\frac{1}{10}\right)^2\right]^{2005}=\left(\frac{1}{10}\right)^{4010}=\frac{1}{10^{4010}}\)
Vậy \(\left(\frac{9}{11}-0,81\right)^{2005}< \frac{1}{10^{4010}}\)
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\(\left(\frac{9}{11}-0,81\right)^{2005}=\left(\frac{9}{1100}\right)^{2005}=0,00\left(81\right)^{2005}\)
\(\frac{1}{10^{4010}}=\frac{1}{100^{2005}}=\left(\frac{1}{100}\right)^{2005}=0,01^{2005}\)
Vì 0,00(81)<0,01 nên \(\left(\frac{9}{11}-0,81\right)^{2005}< \frac{1}{10^{4010}}\)
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a) A = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 + 1/14.17 + 1/17.20
=> 3A = 1/2 - 1/5 + 1/5 - .... + 1/14 - 1/17 + 1/17 - 1/20
=> 3A = 1/2 - 1/20 = 9/20
=> A = 3/20
b) 200410 + 20049 = 20049(1+2004) = 20049 . 2005
200510 = 20059 . 2005
Do 20059 > 20049 nên 200410 + 20049 < 200510
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\(A=\dfrac{5^{10}+1}{5^{11}+1}\)
=>\(5\cdot A=\dfrac{5^{11}+5}{5^{11}+1}=\dfrac{5^{11}+1+4}{5^{11}+1}=1+\dfrac{4}{5^{11}+1}\)
\(B=\dfrac{5^9+1}{5^{10}+1}\)
=>\(5B=\dfrac{5^{10}+5}{5^{10}+1}=1+\dfrac{4}{5^{10}+1}\)
\(5^{11}+1>5^{10}+1\)
=>\(\dfrac{4}{5^{11}+1}< \dfrac{4}{5^{10}+1}\)
=>\(\dfrac{4}{5^{11}+1}+1< \dfrac{4}{5^{10}+1}+1\)
=>5A<5B
=>A<B
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\(A< \frac{\left(10^{10}-1\right)+11}{\left(10^{11}-1\right)+11}< \frac{10^{10}+10}{10^{11}+10}< \frac{10\left(10^9+1\right)}{10\left(10^{10}+1\right)}< \frac{10^9+1}{10^{10}+1}\)
\(\Rightarrow A< B\)
Vậy A<B
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Có : 10A = 10.(10^11-1)/10^12-1 = 10^12-10/10^12-1
Vì : 0 < 10^12-10 < 10^12-1 => 10A < 1 (1)
10B = 10.(10^10+1)/10^11+1 = 10^11+10/10^11+1
Vì : 10^11+10 > 10^11+1 > 0 => 10B > 1 (2)
Từ (1) và (2) => 10A < 10B
=> A < B
Tk mk nha
\(A=\frac{10^{11}-1}{10^{12}-1}\)
\(B=\frac{10^{10}+1}{10^{11}+1}\)
Mà \(\frac{10^{11}-1}{10^{12}-1}< 1\); \(\frac{10^{10}+1}{10^{11}+1}< 1\)
\(\Rightarrow\)\(A,B< 1\)
Ta có:
\(10^{11}-1>10^{10}+1\); \(10^{12}-1>10^{11}+1\)
\(\Rightarrow A>B\)
Vậy A > B