help me!!!

hi

\(A>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)

\(A>\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2015-2014}{2014.2015}\)

\(A>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(A>1-\frac{1}{2015}\)

Mà \(\frac{1}{2015}< \frac{1}{4}\Rightarrow1-\frac{1}{2015}>1-\frac{1}{4}=\frac{3}{4}\Rightarrow A>\frac{3}{4}\)

a, \(A=\frac{1}{2\cdot2}+\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+...+\frac{1}{2011\cdot2011}\)

có :

\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)

...

\(\frac{1}{2011\cdot2011}< \frac{1}{2010\cdot2011}\)

nên :

\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2010\cdot2011}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(\Rightarrow A< 1-\frac{1}{2011}\)

\(\Rightarrow A< \frac{2010}{2011}< 1\)

b, \(A=\frac{2010}{2011}=1-\frac{1}{2011}\) 

\(\frac{3}{4}=1-\frac{1}{4}\)

\(\frac{1}{4}>\frac{1}{2011}\)

nên :

\(A>\frac{3}{4}\)

a, A bé hơn 1

b, A bé hơn 3/4

                                                                     \(Giải\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}\)\(+\frac{1}{4}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2014}\)

      \(A=0+0+0+...+0+0\)

      \(\Rightarrow A=0\)   

\(a.\)\(A< 1\)

b.   \(A< \frac{3}{4}\)

ta co 

1/2.2<1/1*2

...

1/2018*2018<1/2017*2018

=>1/2*2+...+1/2018*1018<1/1*2+...+1/2017.2018

.....(tinh 1/1*2+...+1/2017.*2018)

=>1/2*2+...+1/2018*2018<1-1/2018<1

=>1/2*2+...+1/2018*2018<1

Ta thấy \(\frac{2}{3\times3},\frac{2}{5\times5},\frac{2}{7\times7},\frac{2}{9\times9}>0\)

\(\frac{2}{3\times3}=\frac{2}{9}=\frac{10}{45}>\frac{1}{45}\)

\(\Rightarrow M=\frac{2}{3\times3}+\frac{2}{5\times5}+\frac{2}{7\times7}+\frac{2}{9\times9}>\frac{1}{45}\)

\(\frac{1}{45}\) > m (tổng phép tính)

Ta có: \(\frac{1}{3\times3}<\frac{1}{2\times3}=\frac12-\frac13\)

\(\frac{1}{4\times4}<\frac{1}{3\times4}=\frac13-\frac14\)

...

\(\frac{1}{100\times100}<\frac{1}{99\times100}=\frac{1}{99}-\frac{1}{100}\)

Do đó: \(\frac{1}{3\times3}+\frac{1}{4\times4}+\cdots+\frac{1}{100\times100}<\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{99}-\frac{1}{100}<\frac12\)

=>\(\frac{1}{2\times2}+\frac{1}{3\times3}+\cdots+\frac{1}{100\times100}<\frac{1}{2\times2}+\frac12\)

=>\(P<\frac14+\frac12=\frac34\)

\(A=\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+...+\dfrac{1}{100\times100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Quy đồng 99/100 với 3/4, ta có:

\(\dfrac{99}{100}=\dfrac{396}{400};\dfrac{3}{4}=\dfrac{300}{400}\)

So sánh A với 3/4: \(\dfrac{99}{100}>\dfrac{3}{4}\left(\dfrac{396}{400}>\dfrac{300}{400}\right)\)

\(2m^313dm^3=2,013m^3\)

\(1m^320dm^3=1,02m^3\)

\(4m^35dm^3=4,005m^3\)

\(3m^317cm^3=3,000017m^3\)

2m³ 13dm³ = 3,3m³

1m³ 20dm³ = 3m³

4m³5dm³ = 4,5m³

3m³17cm³ = 3,17m³