A=1/30 + 1/42 + 1/56 + 1/72 + ... + 1/210

A=1/5.6 + 1/6.7 +1/7.8 + 1/8.9 + ... + 1/14.15

A=1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 +...+ 1/14 - 1/15

A=1/5-1/15

A=2/15

a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)

=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)

=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)

=>\(A=\dfrac{127}{128}\)

b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)

29 . ( 85 -47) + 85. ( 47-29 )

=29.38+85.18

=1102+1530

=1632

2,7 . 10,5 - 7,3 . 10,5 - 7,3 . 15 + 2,7 . 15

=(2,7+7,3).10,5-(2,7+7,3).15

=10.10,5-10.15

=105-150

=-45

bạn dùng t/c phân phối là ra mà

phần c tách ra rồi lm thôi cũng dễ mà

Đề sai hả bạn

A=1/6+1/12+1/20+1/30+1/42+1/56+1/72

A=1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9

A=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9

A=1/2-1/9

Câu B tương tự nha bạn :333

1/2-(4/12+9/12)<x<1/24-(3/24-8/24)

1/2-13/12<x<1/24-(-5/24)

-7/12<x<1/4

=>x\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) E{0}

ta có:\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)=\frac{-1}{12}=-0,08333333\)

mà \(\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)=\frac{1}{4}=0.25\)

nên suy ra không có số nguyên x nào thỏa mãn đề bài.

Sao ai cũng giải khó hiểu thế vậy trời:))

ta co

\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}>\frac{1}{2.2}+\frac{1}{3.3}+....+\frac{1}{10.10}\)

ma ve trai =\(1-\frac{1}{10}\)

nen ve phai <1

D=1/1.2+1/2.3+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10

=1+0+0+0+...+0-1/10=1-1/10=9/10

ta có ; 1/2² +1/3²+...+1/2017²<1/1.2+1/2.3+1/3.4+.....+1/2016.2017=1-1/2+1/2-1/3+...+1/2016-1/2017=1+0+0+0+...+0-1/2017

=1-1/2017<1

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2017^2}< \frac{1}{2016.2017}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}=1-\frac{1}{2017}< 1\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< 1\left(đpcm\right)\)

\(A=\dfrac{1}{50}-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=\dfrac{1}{50}-\dfrac{5}{14}\)

\(=\dfrac{14-250}{700}=\dfrac{-236}{700}=-\dfrac{59}{175}\)

Bài 1 : 

Đặt \(A=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\) ta có : 

\(A=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)

\(A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)

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

\(A=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

Vậy \(A< \frac{1}{4}\)

Chúc bạn học tốt ~