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Vì \(B=\frac{2014^{11}+2}{2014^{12}+2}<1\)
\(\Rightarrow B=\frac{2014^{11}+2}{2014^{12}+2}<\frac{2014^{11}+2+4026}{2014^{12}+2+4026}=\frac{2014^{11}+4028}{2014^{12}+4028}=\frac{2014.\left(2014^{10}+2\right)}{2014\left(2014^{11}+2\right)}=\frac{2014^{10}+2}{2014^{11}+2}=A\)
Vậy B<A hay A<B
ta chứng minh bài toán phụ:
nếu ta có b<d \(\frac{a}{b}\)>\(\frac{c}{d}\) thì ad>bc
dễ thây \(\frac{ad}{bd}>\frac{cb}{bd}\)
=> ad>bd
áp dụng:
dat 2014=a ta co
\(A=\frac{a^{10}+2}{a^{11+2}}\)
\(B=\frac{a^{11}+2}{a^{12}+2}\)
ta có
\(A=\frac{a^{10}+2.a^{12}+2}{a^{11}+2.a^{12}+2}\)
\(B=\frac{a^{11}+2.a^{11}+2}{a^{12}+2.a^{11}+2}\)=\(\frac{a^{10}+2a^{12}+2}{a^{12}+2a^{11}+2}\)
=> A=B
mk hok chắc đâu nha
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cùng nhân tử với 2014>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
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\(\frac{2014^{2013}+1}{2014^{2013}-13}\)lớn hơn 1 là \(\frac{14}{2014^{2013}-13}\)
\(\frac{2014^{2012}+8}{2014^{2012}-11}\)lớn hơn 1 là \(\frac{19}{2014^{2012}-11}\)
\(\frac{14}{2014^{2013}-13}\)\(< \)\(\frac{19}{2014^{2012}-11}\)
\(\Rightarrow A< B\)
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Ta có : A = \(\frac{2^{2014}+1}{2^{2014}}=1+\frac{1}{2^{2014}}\)
B = \(\frac{2^{2014}+2}{2^{2014}+1}=1+\frac{1}{2^{2014}+1}\)
Vì : \(\frac{1}{2^{2014}}>\frac{1}{2^{2014}+1}\)
Nên A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
Gợi ý nhé: bạn hãy so sánh 2014A và 2014B rồi suy ngược lại A và B
Ta có:
2014A=20142014+ 2014/20142014+1=1+2013/20142014+1
2014B=20142013+2014/20142013+1=1+2013/20142013+1
vì 1+2013/20142014+1<1+2013/20142013+1 nên 10A < 10B
suy ra A<B
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ta thấy:
2^2014<2^2014+2
=>\(\frac{2^{2014}+1}{2^{2014}}>\frac{2^{2014}+1}{2^{2014}+2}\)
vậy......
Có : 22014 + 1 > 22014 nên \(\frac{2^{2014}+1}{2^{2014}}\)> 1 .
22104 + 1 < 22014 + 2 nên \(\frac{2^{2014}+1}{2^{2014}+2}\)< 1.
=> \(\frac{2^{2014}+1}{2^{2014}}\)>\(\frac{2^{2014}+1}{2^{2014}+2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt :
\(A=\frac{2^{2014}+1}{2^{2014}}=\frac{2^{2014}}{2^{2014}}+\frac{1}{2^{2014}}=1+\frac{1}{2^{2014}}\)
\(B=\frac{2^{2014}+2}{2^{2014}+1}=\frac{2^{2014}+1+1}{2^{2014}+1}=\frac{2^{2014}+1}{2^{2014}+1}\)\(=1+\frac{1}{2^{2014}+1}\)
\(1+\frac{1}{2^{2014}}>1+\frac{1}{2^{2014}+2}\Leftrightarrow A>B\)
B = 201410+2/201411+2 < 201411+2+4026 / 201412+2+4026
= 201411+4028/201412+4028
= 2014(201410+2)/2014(201411+2)
= 201410+2/201411+2 = A
=> A > B