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Gọi A=\(\frac{989898.89-898989.98}{2^3+3^4+...+2014^{2015}}\)
A=\(\frac{98.10001.89-89.10001.98}{2^3+3^4+...+2014^{2015}}\)
A=\(\frac{98.89.\left(10001-10001\right)}{2^3+3^4+...+2014^{2015}}\)
A=\(\frac{98.89.0}{2^3+3^4+...+2014^{2015}}\)
A=\(\frac{0}{2^3+3^4+...+2014^{2015}}\)
A=0
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a) 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10 + 11
= (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + (9 - 10) + 11
= -1 + (-1) + (-1) + (-1) + (-1) + 11
= -1 x 5 + 11
= -5 + 11
= 6
b) 98 + 87 + 76 + ... + 21 - 12 - 23 - 34 - ... - 89
= (98 89) + (87 - 78) + ... + (21 - 12) (có 8 nhóm)
= 9 + 9 + ... + 9 (có 8 số 9)
= 9 x 8
= 72
\(1-2+3-4+.....-10+11\)
\(=\left(1-2\right)+\left(3-4\right)+...\left(9-10\right)+11\)
\(=1+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+11\)
\(=-1\times5+11\)
\(=6\)
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a) 5 13 + − 5 7 + − 20 41 + 8 13 + − 21 41 = 5 13 + 8 13 + − 20 41 + − 21 41 + − 5 7 = − 5 7
b) 1 28 + − 1 14 + 3 28 + − 1 7 + 3 14 = 1 28 + 3 28 + − 1 14 + 3 14 + − 1 7 = 1 7
c) 1 2 + − 2 3 + 3 4 + − 4 5 + 5 6 + − 5 6 + 4 5 + − 3 4 + 2 3 + − 1 2 = 1 2 + − 1 2 + 3 4 + − 3 4 + − 4 5 + 4 5 + 5 6 + − 5 6 = 0
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1 2 + − 1 2 + − 2 3 + 2 3 + 3 4 + − 3 4 + − 4 5 + 4 5 + 5 6 + − 5 6 = 0
1 2 + − 1 2 + − 2 3 + 2 3 + 3 4 + − 3 4 + − 4 5 + 4 5 + 5 6 + − 5 6 = 0
Đây nha
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\(A=\frac{98^{2015}+1}{98^{2014}+1}>1\)
Ta có:
\(A=\frac{98^{2015}+1+97}{98^{2014}+1+97}=\frac{98^{2015}+98}{98^{2014}+98}=\frac{98\left(98^{2014}+1\right)}{98\left(98^{2013}+1\right)}\)
\(=\frac{98\left(98^{2015}+1\right)}{98\left(98^{2014}+1\right)}=\frac{98^{2014}+1}{98^{2013}+1}\)
Ta thấy: \(\frac{98^{2014}+1}{98^{2013}+1}=B\)mà \(A>1\)
\(\Rightarrow A>B\)
\(A=\frac{98^{2015}+1}{98^{2014}+1}>1\)
Theo đề ta có:
\(A=\frac{98^{2015}+1+97}{98^{2014}+1+97}=\frac{98^{2015}+98}{98^{2014}+98}=\frac{98\left(98^{2014}+1\right)}{98\left(98^{2013}+1\right)}\)
\(=\frac{98\left(98^{2015}+1\right)}{98\left(98^{2014}+1\right)}=\frac{98^{2014}+1}{98^{2013}+1}\)
Lúc này ta thấy: \(\frac{98^{2014}+1}{98^{2013}+1}=B\)mà \(A>1\)
\(\Leftrightarrow A>B\).