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(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)
\(\frac{1234\times1235-1}{1234\times1235}=1-\frac{1}{1234\times1235};\frac{1235\times1236-1}{1235\times1236}=1-\frac{1}{1235\times1236}=1-\frac{1}{1235\times1234+1235\times2}\)
Vì 1234*1235 < 1235*1234+1235*2 nên \(\frac{1}{1234\times1235}>\frac{1}{1235\times1234+1235\times2}\).
Do đó \(1-\frac{1}{1234\times1235}<1-\frac{1}{1235\times1234+1235\times2}\)
Vậy \(\frac{1234.1235-1}{1234.1235}<\frac{1235.1236-1}{1235.1236}\)
\(\frac{1234\times1235-1}{1234\times1235}\&\frac{1235\times1236-1}{1235\times1236}\)
Đề vậy hả
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 \(\frac{1}{1234.1235}>\frac{1}{1235.1236}\)
\(\Rightarrow\frac{1234.1235-1}{1234.1235}< \frac{1235.1236-1}{1235.1236}\)
a) Ta có: a < b => a + 1 < b + 1
b) Ta có: a < b => a - 2 < b - 2
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot...\left(\frac{1}{10}-1\right)\)
\(A=\left(\frac{1}{2}-\frac{2}{2}\right)\left(\frac{1}{3}-\frac{3}{3}\right)\cdot...\cdot\left(\frac{1}{10}-\frac{10}{10}\right)\)
\(A=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot...\cdot\left(-\frac{9}{10}\right)\)
\(A=\frac{-1}{2}\cdot\frac{-2}{3}\cdot...\cdot\frac{-9}{10}\)
\(A=\frac{\left(-1\right)\cdot\left(-2\right)\cdot...\cdot\left(-9\right)}{2\cdot3\cdot...\cdot10}\)
\(A=\frac{\left(-1\right)\cdot2\cdot...\cdot9}{2\cdot3\cdot...\cdot10}=\frac{-1}{10}\)
Mà \(\frac{-1}{10}>\frac{-1}{9}\)nên A > -1/9
Phần cuối tương tự
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 \(\frac{1}{1234.1235}>\frac{1}{1235.1236}\)
=> \(\frac{1234.1235-1}{1234.1235}>\frac{1235.1236-1}{1235.1236}\)