! là j z

\("!"\)  là giai thừa đó bạn ạ .

\(VD:\)  \(3!=1.2.3=6\)

          \(4!=1.2.3.4=24\)

A=1/2*3/4*..*99/100

=>A<2/3*4/5*6/7*...*100/101

=>A^2<2/3*4/5*...*100/101*1/2*3/4*...*99/100

=>A^2<1/101<1/100

=>A<1/10

 \(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\)

\(A< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\right)\)

\(\Rightarrow A=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{98}{99}.\dfrac{99}{100}\)

\(\Leftrightarrow A=\dfrac{1.2.3.4.5.6...98.99}{2.3.4.5.6.7...99.100}\)

\(\Rightarrow A< \dfrac{1}{100}< \dfrac{1}{10}\)

Vậy \(A< \dfrac{1}{10}\)

a)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)

=\(\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{100}{100!}-\frac{1}{100!}\)

=\(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)

=\(1-\frac{1}{100!}< 1\)

\(\Rightarrow\)\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)

b)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

=\(\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

=\(\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\right)\)=\(1+1-\frac{1}{99}-\frac{1}{100}\)

=\(2-\frac{1}{99}-\frac{1}{100}< 2\)

\(\Rightarrow\)\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)

bgcfd

đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{98}{99}.\frac{100}{100}\Leftrightarrow A

Từ 1->100 có:100-1+1=100 (thừa số)

Mà \(\frac{1}{2};\frac{3}{4};\frac{5}{6};.....;\frac{99}{100}\) là những p/s có tử và mẫu là 2 số liên tiếp

=>từ \(\frac{1}{2}\rightarrow\frac{99}{100}\) có : 50 thừa số

=>M có 50 thừa số

Từ 2->101 có:101-2+1=100 (thừa số)

=>từ \(\frac{2}{3}\rightarrow\frac{100}{101}\) có: 50 thừa số

=>N có 50 thừa số

Do đó mỗi biểu thức M,N đều có 50 thừa số

Mà \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};......;\frac{99}{100}< \frac{100}{101}\)

=>\(M=\frac{1}{2}.\frac{2}{3}.......\frac{99}{100}< N=\frac{2}{3}.\frac{4}{5}.........\frac{100}{101}\)

Vậy M<N

Chả hiểu đề ra làm sao 

so sanh 1/2 o m voi 100/101(1/2<100/101)

nhe k nhe ban than

đường cong parabol