Ta có :

\(\frac{1}{101}>\frac{1}{200}\)

\(\frac{1}{102}>\frac{1}{200}\)

\(\frac{1}{103}>\frac{1}{200}\)

\(.........\)

\(\frac{1}{200}=\frac{1}{200}\)

Cộng vế với vế ta được :

\(\frac{1}{101}+\frac{1}{102}+.....+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+....+\frac{1}{200}\) (có 100 số hạng \(\frac{1}{200}\))\(=\frac{100}{200}=\frac{1}{2}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{200}>\frac{1}{2}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{204}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{203}{204}\)

\(=\frac{1}{204}\)

\(\text{Sửa đề }\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times....\times\left(1-\frac{1}{203}\right)\times\left(1-\frac{1}{204}\right)\)

\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times....\times\frac{202}{203}\times\frac{203}{204}\)

\(=\frac{1\times2\times3\times...\times202\times203}{2\times3\times4\times...\times203\times204}\)

\(=\frac{1}{204}\)

\(\left(1-\frac{1}{2}\right)\)\(.\left(1-\frac{1}{3}\right)\)\(...\left(1-\frac{1}{203}\right).\left(1-\frac{1}{204}\right)\)

\(=\frac{1}{2}.\frac{2}{3}....\frac{202}{203}.\frac{203}{204}\)

=\(\frac{1.2....202.203}{2.3....203.204}\)

=\(\frac{1}{204}\)

Lạy ông đi qua,bà đi lại.

Cho con một tích đúng

ai nhanh nhất mình tk cho

cămmon

1, A=1/2 x (1/2 + 1/6 + 1/12+ 1/20 +1/30 +1/42)

    A= 1/2 x (1/1x2 + 1/ 2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7)

A=1/2 x (1/1 -1/2 +1/2 -1/3 +1/3 - 1/4 + 1/4 -1/5 + 1/5 - 1/6 +1/6 - 1/7 )

A= 1/2 x (1- 1/7)

A= 1/2 x 6/7

A=3/7

\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}\)

\(\dfrac{1}{2.2}< \dfrac{1}{1.2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3.3}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{4.4}< \dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)

...

\(\dfrac{1}{2009.2009}< \dfrac{1}{2008.2009}=\dfrac{1}{2008}-\dfrac{1}{2009}\)

\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...\dfrac{1}{2008}-\dfrac{1}{2009}=1-\dfrac{1}{2009}< 1\)

\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}< 1\)

Ta có:

\(\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+\dfrac{1}{4\times4}+...+\dfrac{1}{2009\times2009}< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{2008\times2009}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}=1-\dfrac{1}{2009}< 1\)

=(1-2)-(3-4)+(5-6)-(7-8)+...+(2021-2022)-2023
=(-1)-(-1)+(-1)-...+(-1)-2023
=0-2023
=-2023

OA là tia nằm giữa hai tia OC và OB

OA là tia nằm giữa hai tia OC và OB

OA là tia phân giác của BOC

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

= \(\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{12}+\frac{1}{24}\right)+\left(\frac{1}{48}+\frac{1}{96}\right)+\frac{1}{192}\)

=               \(\left(\frac{1}{2}+\frac{1}{8}\right)+\left(\frac{1}{32}+\frac{1}{192}\right)\)

=                              \(\frac{5}{8}+\frac{1}{192}\)

=                                    \(\frac{121}{192}\)

ko ai bết đâu

A=1-2-3+4+5-6-7+8+...+97-98-99+100

=>A=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)

=>A=0+0+....+0=0

vậy A=0

B=1-2+2^2-2^3+...+2^100

=>2B=2-2^2+2^3-2^4+....+2^101

=>2B+B=1-2^101=3B

=>B=1-2^101/3

C= 2^100-2^99-2^98-...-2^2-2-1

=>C=2^100-(2^99+2^98+.....+2^2+2+1)

Đặt D=2^99+2^98+.....+2^2+2+1

=>2D=2^100+2^99+.....+2^3+2^2+2

=>2D-D=2^100-1=D

=>C=2^100-(2^100-1)=1

tick nha

hic!ngày kia phải nộp rồi ! mọi người giúp mình nhanh nha!

1+2-3-4+.....+97+98-99-100 nhanh len