Ý gì vậy

Sai là 0 👈👈👈👈👈👈👈👈👈👈👈

3) Ta có : \(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

4)

A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

A = \(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{99}-\frac{1}{101}\right)\)

A = \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

A = \(\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(A=\frac{1}{2}.\frac{100}{101}\)

A = \(\frac{50}{101}\)

2, đặt tên biểu thức trên là A. Ta có :

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(A=1-\frac{1}{101}\)

\(A=\frac{100}{101}\)

1) \(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)

\(=1-\frac{1}{5}\)

\(=\frac{4}{5}\)

@Ace Legona

Nhận thấy \(\)\(\dfrac{1}{1.1!}=1\); \(\dfrac{1}{2.2!}=\dfrac{1}{4}\)

Đặt \(P=\dfrac{1}{3.3!}+...+\dfrac{1}{2013.2013!}\)

\(P=\dfrac{1}{3.1.2.3}+...+\dfrac{1}{2013.1.2...2013}\)

\(P< \dfrac{1}{1.2.3}+...+\dfrac{1}{2011.2012.2013}\)

\(P< \dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}-\dfrac{1}{2012.2013}\right)\)

\(P< \dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2012.2013}\right)=\dfrac{1}{4}-\dfrac{1}{2.2012.2013}\)

\(P< \dfrac{1}{4}\)

\(A< \dfrac{1}{4}+\dfrac{1}{4}+1=\dfrac{3}{2}\left(đpcm\right)\)

Xàm hả!!!!!!!!!

toán j lạ vậy

toán đúng rồi đó ban, nhưng mình làm rồi

\(\frac{1}{1.1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{49.25}\) 

\(=\frac{2}{2}.\left(\frac{1}{1.1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{49.25}\right)\)

\(=\frac{2}{1.1.2}+\frac{2}{1.3.2}+\frac{2}{3.2.2}+...+\frac{2}{49.25.2}\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=2.\left(1-\frac{1}{50}\right)\)

\(=2.\frac{49}{50}\)

\(=\frac{49}{25}\)

Chúc bạn học tốt !!! 

Bài 2:

100! = 1.2.3.4.   ...................   .100

Thấy :            \(\frac{1}{1.1!}=\frac{1}{1}\)

                       \(\frac{1}{2.2!}=\frac{1}{4}\)

                       \(\frac{1}{3.3!}< \frac{1}{1.2.3}\)( Vì 3.3! > 1.2.3 )

                         ...

                       \(\frac{1}{2019.2019!}< \frac{1}{2017.2018.2019}\)( vì 2019.2019! < 2017.2018.2019)

Cộng từng vế có :

  \(\frac{1}{3.3!}+\frac{1}{4.4!}+...+\frac{1}{2019.2019!}< \frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)

\(\Rightarrow\frac{1}{1.1!}+\frac{1}{2.2!}+...+\frac{1}{2019.2019!}< \frac{1}{1}+\frac{1}{4}+\frac{1}{1.2.3}+...+\frac{1}{2017.2018.2019}\)

\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right):2\)

\(\Rightarrow C< \frac{1}{1}+\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2018.2019}\right):2\)

\(\Rightarrow C< \frac{3}{2}-\frac{1}{2.2018.2019}\)

Vì \(\frac{1}{2.2018.2019}>0\Rightarrow C< \frac{3}{2}\)

\(\text{Đặt biểu thức là A:}\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}+\frac{1}{100^2}\)

\(\text{Ta có:}\frac{1}{2^2}=\frac{1}{2\times2}< \frac{1}{1\times2}\)

\(\frac{1}{3^2}=\frac{1}{3\times3}< \frac{1}{2\times3}\)

\(\frac{1}{4^2}=\frac{1}{4\times4}< \frac{1}{3\times4}\)

\(...\)

\(\frac{1}{99^2}=\frac{1}{99\times99}< \frac{1}{98\times99}\)

\(\frac{1}{100^2}=\frac{1}{100\times100}=\frac{1}{99\times100}\)

\(\Rightarrow A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< 1-\frac{1}{100}< 1\)

\(\Rightarrow A< 1\left(đpcm\right)\)