Đề có phải thế này không nếu thế thì mình chiều 5 giờ mình giải cho giờ mình mắc rồi:

Cho B=7+7³+7⁵+7⁷+ .........+7⁹⁹.Chứng tỏ B chia hết cho 48

Mình hứa

Vào bảng chat đi mình nhắn đó

câu a
\(A=\frac{33.10^3}{2^3.5.10^3+7000}=\frac{33.10^3}{2^3.5.10^3+7.10^3}=\frac{33.10^3}{10^3\left(2^3.5+7\right)}=\frac{33.10^3}{10^3.47}=\frac{33}{47}\)
\(B=\frac{3774}{5217}=\frac{34.111}{47.111}=\frac{34}{47}\)
\(\Rightarrow\frac{33}{47}< \frac{34}{47}\)
=> A<B

\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(C=7.\frac{3}{35}\)

\(C=\frac{3}{5}\)

chắc chưa tính lại thử đi

Giải:

a) \(A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{-20}{41}+\dfrac{5}{13}+\dfrac{-21}{41}\)

\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{5}{13}+\dfrac{-21}{41}+\dfrac{-20}{41}\)

\(\Leftrightarrow A=\dfrac{5}{13}\left(\dfrac{5}{7}+1\right)+\dfrac{-41}{41}\)

\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{12}{7}+\left(-1\right)\)

\(\Leftrightarrow A=\dfrac{60}{91}+\left(-1\right)=-\dfrac{31}{91}\)

Vậy ...

b) \(B=\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{12}{11}-\dfrac{5}{7}.\dfrac{7}{11}\)

\(\Leftrightarrow B=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{12}{11}-\dfrac{7}{11}\right)\)

\(\Leftrightarrow B=\dfrac{5}{7}.\dfrac{7}{11}\)

\(\Leftrightarrow B=\dfrac{5}{11}\)

Vậy ...

c) \(C=\dfrac{-2}{3}+\dfrac{-5}{7}+\dfrac{2}{3}+\dfrac{-2}{7}\)

\(\Leftrightarrow C=\left(\dfrac{-2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-2}{7}+\dfrac{-5}{7}\right)\)

\(\Leftrightarrow C=0+\left(-1\right)=-1\)

Vậy ...

A

Ta có A= -1+7+(-7²)+7³+(-7⁴)+....+7²⁰⁰⁸ +7²⁰⁰⁸

        A.7=[-7+7²+(-7³⁾+7⁴+....+7²⁰⁰⁹ +7^{2009]  + [} -1+7+(-7²)+7³+(-7⁴)+....+7²⁰⁰⁸ +7^2008]

        A.7=[7²⁰⁰⁹.2+(-1) +7^2008] :7

b;c làm tương tự

329534849

\(=7\left(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{69.70}\right)\)

\(=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(=7\left(\frac{1}{10}-\frac{1}{70}\right)\)

\(=7\times\frac{6}{70}\)

\(=\frac{6}{10}=\frac{3}{5}\)

Gọi tổng trên là A

A = 7/10.11 + 7/11.12 +.....+ 7/69.70

A = 7(1/10.11 + 1/11.12 +.....+ 1/69.70)

A =7( 1/10 - 1/11 + 1/11 - 1/12 +.....+ 1/69 - 1/70)

A = 7( 1/10 - 1/70)

A = 7 . 3/35

A = 21/35

\(a)\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)

\(=\dfrac{13}{4}-\dfrac{7}{-24}\)

\(=\dfrac{85}{24}\)

\(b)\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)

\(=\dfrac{-3}{56}-\dfrac{3}{28}\)

\(=\dfrac{-9}{56}\)

\(c)\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}\)

\(=\dfrac{13}{12}\)\(+\dfrac{-2}{3}\)

\(=\dfrac{5}{12}\)

\(d)\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-1}{14}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-23}{126}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-4}{7}+\dfrac{4}{7}\)

\(=0\)

\(e)\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{-5}{56}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{83}{56}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{305}{168}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\dfrac{47}{24}+\dfrac{5}{-8}\)

\(=\dfrac{4}{3}\)

Bài 2 : Tính

a) \(\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)

\(=\dfrac{18}{24}-\dfrac{-60}{24}-\dfrac{-4}{24}\)

\(=\dfrac{18-\left(-60\right)-\left(-7\right)}{24}\)

\(=\dfrac{85}{24}\)

b) \(\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)

\(=\dfrac{32}{56}+\dfrac{-35}{56}-\dfrac{6}{56}\)

\(=\dfrac{32+\left(-35\right)-6}{56}\)

\(=\dfrac{-9}{56}\)

c) \(\dfrac{7}{36}-\dfrac{8}{9}+\dfrac{-2}{3}\)

\(=\dfrac{7}{36}-\dfrac{32}{36}+\dfrac{-24}{36}\)

\(=\dfrac{7-32+\left(-24\right)}{36}\)

\(=\dfrac{-49}{36}\)

d) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\dfrac{-9}{18}+\dfrac{3}{7}-\dfrac{2}{18}+\dfrac{-7}{18}+\dfrac{4}{7}\)

\(=\left(\dfrac{-9}{18}+\dfrac{-7}{18}-\dfrac{2}{18}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=\left(-1\right)+1\)

\(=0\)

e) \(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)

\(=\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{11}{7}\right)+\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\dfrac{1}{3}\)

\(=2+\left(-1\right)+\dfrac{1}{3}\)

\(=1+\dfrac{1}{3}\)

\(=\dfrac{4}{3}\)

\(a⋮7\);  \(b⋮7\)

\(a,b\in N\). a , b có thể là bất cứ một số nào miễn sao chia hết cho 7.

Với việc x , y \(\in\)N,ax\(⋮\)7 và by\(⋮\)7 thì tức là a và b được nhân lên nhiều lần và ĐK \(⋮\)7 không hề bị thay đổi.

\(\Rightarrow ax,by⋮7\)