2^2016 <7^730

10^201  < 90^101

cảm ơn bạn mình cũng đang chuẩn bị đăng câu hỏi này nè

Ta có: \(\frac{-11}{13}=\frac{-13+2}{13}=-1+\frac{2}{13}\)

\(\frac{99}{-101}=\frac{-99}{101}=\frac{-101+2}{101}=-1+\frac{2}{101}\)

\(\frac{-199}{201}=\frac{-201+2}{201}=-1+\frac{2}{201}\)

mà \(\frac{2}{13}>\frac{2}{101}>\frac{2}{201}\) (13<101<201)

nên \(\frac{-11}{13}>\frac{99}{-101}>\frac{-199}{201}\)

A=2016^2016+2/2016^2016-1>1

=>(2016^2016)+2/(2016^2016)-1<(2016^2016)+2-2/(2016^2016)-1-2=2016^2016/(2016^2016)-3=B

> vì pst1 >1 pst2<1

\(>\)

So sánh:

\(99 \times 201 \text{v}ớ\text{i} 199 \times 101\)

• \(99 \times 201 = 99 \times \left(\right. 200 + 1 \left.\right) = 99 \times 200 + 99 = 19800 + 99 = 19899\)
• \(199 \times 101 = 199 \times \left(\right. 100 + 1 \left.\right) = 19900 + 199 = 20099\)

So sánh:

\(19899 < 20099\)

→ \(\frac{99}{101} < \frac{199}{201}\)

Kết quả:

\(\frac{99}{101} < \frac{199}{201}\)

\(\frac{99}{101}và\frac{199}{201}\)

\(\frac{99}{101}=1-\frac{2}{101};\frac{199}{201}=1-\frac{2}{201}\)

\(\) Vì \(\frac{2}{101}>\frac{2}{201}\) nên \(1-\frac{2}{101}<1-\frac{2}{201}\)

Vậy \(\frac{99}{101}<\frac{199}{201}\)