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b) Ta có:
\(\frac{1234.1235-1}{1234.1235}=1-\frac{1}{1234.1235}\)
\(\frac{1235.1236-1}{1235.1236}=1-\frac{1}{1235.1236}\)
DO \(\frac{1}{1234.1235}>\frac{1}{1235.1236}\)=> \(-\frac{1}{1234.1235}< -\frac{1}{1235.1236}\)
=> \(\frac{1234.1235-1}{1234.1235}< \frac{1235.1236-1}{1235.1236}\)
a) Ta có:
\(-\frac{31}{32}< 0< \frac{31317}{32327}\)
b) Ta có:
\(1-\frac{1234.1235-1}{1234.1235}=\frac{1}{1234.1235}\)
\(1-\frac{1235.1236-1}{1235.1236}=\frac{1}{1235.1236}\)
Mà \(\frac{1}{1234.1235}>\frac{1}{1235.1236}\)
\(\Rightarrow\frac{1234.1235-1}{1234.1235}< \frac{1235.1236-1}{1235.1236}\)
\(\frac{1234\times1235-1}{1234\times1235}=1-\frac{1}{1234\times1235};\frac{1235\times1236-1}{1235\times1236}=1-\frac{1}{1235\times1236}=1-\frac{1}{1235\times1234+1235\times2}\)
Vì 1234*1235 < 1235*1234+1235*2 nên \(\frac{1}{1234\times1235}>\frac{1}{1235\times1234+1235\times2}\).
Do đó \(1-\frac{1}{1234\times1235}
\(\frac{1234\times1235-1}{1234\times1235}\&\frac{1235\times1236-1}{1235\times1236}\)
Đề vậy hả
(1234.1235-1)/(1234.1235)=1-1/(1234.1235)
(1235.1236-1)/(1235.1236)=1-1/(1235.1236)
Vì 1/(1234.1235)>1/(1235.1236) nên 1-1/(1234.1235)<1-1/(1235.1236) hay (1234.1235-1)/(1234.1235) < (1235.1236-1)/(1235.1236)
Ta có:
\(\frac{1234.1235-1}{1234.1235}=\frac{1234.1235}{1234.1235}-\frac{1}{1234.1235}=1-\frac{1}{1234.1235}\)
\(\frac{1235.1236-1}{1235.1236}=\frac{1235.1236}{1235.1236}-\frac{1}{1235.1236}=1-\frac{1}{1235.1236}\)
Vì 1/1234.1235 > 1/1235.1236
=> 1 - 1/1234.1235 < 1 - 1/1235.1236
Vậy .....
Ta có: \(\frac{1234.1235-1}{1234.1235}=\frac{1234.1235}{1234.1235}-\frac{1}{1234.1235}\)
= \(1-\frac{1}{1234.1235}\)
Lại có: \(\frac{1235.1236-1}{1235.1236}=\frac{1235.1236}{1235.1236}-\frac{1}{1235.1236}\)
= \(1-\frac{1}{1235.1236}\)
Vì 1234.1235 < 1235.1236 nên \(\frac{1}{1234.1235}>\frac{1}{1235.1236}\)
=> \(\frac{1234.1235-1}{1234.1235}>\frac{1235.1236-1}{1235.1236}\)
31/32= 1 -1/32
31317/32327= 1- 1010/32327
ta có 1010/32327 x 32 = 32320/32327 <1=32/32=1/32 x 32
=> 1010/32327<1/32 => 31317/32327 > 31/32
Ta có\(\frac{-31}{-32}=\frac{31}{32}=\frac{31310}{32320}\)
Vì 31310<32320 nên \(\frac{31310}{32320}< 1\)
\(\Rightarrow\frac{31310}{32320}< \frac{31310+7}{32320+7}=\frac{31317}{32327}\)
\(\frac{-31}{-32}< \frac{31317}{32327}\)
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