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\(\frac{200}{201}+\frac{201}{202}>\frac{200}{201+202}+\frac{201}{201+202}=\frac{200+201}{201+202}\)
\(A=\frac{54.107-53}{53.107+54}=\frac{\left(53+1\right)107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{\left(134+1\right)269-133}{134.269+135}=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
Vậy A<B
câu a là
200/201+201/202>200/202+201/202>1
200+201/201+201<1
=>200/201+201/202>1>200+201/201+202
A=\(\frac{54.107-53}{53.107+54}\)
= \(\frac{54.\left(106+1\right)-53}{53.107+53+1}\)
= \(\frac{54.106+54-53}{53.\left(107+1\right)+1}\)
= \(\frac{54.106+1}{53.108+1}\)
mà \(54.106=53.108\)
=> \(\frac{54.106+1}{53.108+1}=1\)
B = \(\frac{135.269-133}{134.269+135}\)
= \(\frac{135.\left(268+1\right)-133}{134.269+134+1}\)
= \(\frac{135.268+135-133}{134.\left(269+1\right)+1}\)
= \(\frac{135.268+2}{134.270+1}\)
Mà \(135.268=134.270\)
=> \(\frac{135.268+2}{134.270+1}=\frac{135.268+1+1}{134.270+1}\)
\(=\frac{135.268+1}{134.270+1}+\frac{1}{134.270+1}\)
\(=1+\frac{1}{134.270+1}>1\)
=> B > 1
=> B > A
A= (54.107-53)/(53.107+54)
= (53+1).107-53 / 53.107+54
=53.107+107-53 / 53.107+54
=53.107+54 / 54.107 + 54
=1
B= 135.269-133 / 134.269+135
= (134+1).269-133 / 134.269+135
= 134.269+269-133 / 134.269+135
=134.269+136 / 134.269+135
=134.269+135/ 134.269+135 + 1/134.269+135
=1 + 1/134.269+135 >1=A
\(A=\frac{54.107-53}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}>1\)
\(A=\frac{54.107-53}{53.107+54}<\frac{135.269-133}{134.269+135}\)
THAM KHẢO :
A = 53.107 + 54 53 + 1 .107 − 53
= 53.107 + 54 53.107 + 107 − 53
= 53.107 + 54 53.107 + 54 = 1.
B = 134.269 + 135 134 + 1 .269 − 133
= 134.269 + 135 134.269 + 269 − 133
= 134.269 + 135 134.269 + 136 > 1
⇒A < B
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