Bài 10:

$-A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}$

$=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}$

$=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+...+\frac{10-9}{9.10}$

$=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}$

$=\frac{1}{4}-\frac{1}{10}=\frac{3}{20}$

$\Rightarrow A=\frac{-3}{20}$

Bài 11:

$A=\frac{2n}{n+3}=\frac{2(n+3)-6}{n+3}=2-\frac{6}{n+3}$
Để $A$ nguyên thì $\frac{6}{n+3}$ nguyên.

Với $n$ nguyên thì điều trên xảy ra khi $6\vdots n+3$

$\Rightarrow n+3\in\left\{\pm 1; \pm 2; \pm 3; \pm 6\right\}$

$\Rightarrow n\in\left\{-4; -2; -1; -5; -6; 0; -9; 3\right\}$

Ta có : 2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-82x+2x+1+2x+2+...+2x+2015=22019−8

\Leftrightarrow2^x\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-8⇔2x(1+2+22+...+22015)=22019−8 (1)

Đặt : A=1+2+2^2+...+2^{2015}A=1+2+22+...+22015

\Rightarrow2A=2+2^2+2^3+...+2^{2016}⇒2A=2+22+23+...+22016

\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)⇒2A−A=(2+22+23+...+22016<...

$2^x+2^{x+1}+...+2^{x+2015}=2^{2019}-8$

$\to 2^x.1+2^x.2+....+2^x.2^{2015}=2^{2019}-8$

$\to 2^x.(1+2+...+2^{2015})=2^{2019}-8$

Đặt:

$A=1+2+...+2^{2015}$

$2A=2.(1+2+...+2^{}$

6 đội,16 người

Bài 1

4n - 6 = 4n - 2 - 4 = 2(2n - 1) - 4

Để (4n - 6) ⋮ (2n - 1) thì 4 ⋮ (2n - 1)

⇒ 2n - 1 ∈ Ư(4) = {-4; -2; -1; 1; 2; 4}

⇒ 2n ∈ {-3; -1; 0; 2; 3; 5}

⇒ n ∈ {-3/2; -1/2; 0; 1; 3/2; 5/2}

Mà n là số tự nhiên

⇒ n ∈ {0; 1}

Bạn Kiều Vũ Linh cho mình hỏi là 3/2 là phân số hả bạn ??

Bài 57:

\(A=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\cdots+\frac{2}{97\cdot100}\)

\(=\frac23\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\cdots+\frac{3}{97\cdot100}\right)\)

\(=\frac23\left(1-\frac14+\frac14-\frac17+\cdots+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac23\left(1-\frac{1}{100}\right)=\frac23\cdot\frac{99}{100}=\frac{33}{50}\)

Bài 52:

\(A=\frac{4}{7\cdot31}+\frac{6}{7\cdot41}+\frac{9}{10\cdot41}+\frac{7}{10\cdot57}\)

\(=\frac{20}{31\cdot35}+\frac{30}{35\cdot41}+\frac{45}{41\cdot50}+\frac{35}{50\cdot57}\)

\(=5\left(\frac{4}{31\cdot35}+\frac{6}{35\cdot41}+\frac{9}{41\cdot50}+\frac{7}{50\cdot57}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)\)

\(=5\left(\frac{1}{31}-\frac{1}{57}\right)\)

\(B=\frac{7}{19\cdot31}+\frac{5}{19\cdot43}+\frac{3}{23\cdot43}+\frac{11}{23\cdot57}\)

\(=\frac{14}{31\cdot38}+\frac{10}{38\cdot43}+\frac{6}{43\cdot46}+\frac{22}{46\cdot57}\)

\(=2\left(\frac{7}{31\cdot38}+\frac{5}{38\cdot43}+\frac{3}{43\cdot46}+\frac{11}{46\cdot57}\right)\)

\(=2\left(\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}\right)=2\left(\frac{1}{31}-\frac{1}{57}\right)\)

Do đó: \(\frac{A}{B}=\frac52\)

Bài 53:

\(A=\frac{34}{7\cdot13}+\frac{51}{13\cdot22}+\frac{85}{22\cdot37}+\frac{68}{37\cdot49}\)

\(=17\cdot\left(\frac{2}{7\cdot13}+\frac{3}{13\cdot22}+\frac{5}{22\cdot37}+\frac{4}{37\cdot49}\right)\)

\(=\frac{17}{3}\left(\frac{6}{7\cdot13}+\frac{9}{13\cdot22}+\frac{15}{22\cdot37}+\frac{12}{37\cdot49}\right)\)

\(=\frac{17}{3}\left(\frac17-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)=\frac{17}{3}\left(\frac17-\frac{1}{49}\right)\)

\(B=\frac{39}{7\cdot16}+\frac{65}{16\cdot31}+\frac{52}{31\cdot43}+\frac{26}{43\cdot49}\)

\(=13\left(\frac{3}{7\cdot16}+\frac{5}{16\cdot31}+\frac{4}{31\cdot43}+\frac{2}{43\cdot49}\right)\)

\(=\frac{13}{3}\left(\frac{9}{7\cdot16}+\frac{15}{16\cdot31}+\frac{12}{31\cdot43}+\frac{6}{43\cdot49}\right)\)

\(=\frac{13}{3}\left(\frac17-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)=\frac{13}{3}\left(\frac17-\frac{1}{49}\right)\)

Do đó: \(\frac{A}{B}=\frac{17}{3}:\frac{13}{3}=\frac{17}{13}\)

Bài khó lắmmm tr oiii

B = \(\dfrac{4}{7}\) = \(\dfrac{4.4}{7.4}\) = \(\dfrac{16}{28}\); I = \(\dfrac{6}{13}\) = \(\dfrac{6.\left(-2\right)}{13.\left(-2\right)}\) = \(\dfrac{-12}{-26}\)

N = \(\dfrac{-5}{13}\) = \(\dfrac{-5.3}{13.3}\) = \(\dfrac{-15}{39}\); T = \(\dfrac{7}{21}\) = \(\dfrac{7.4}{21.4}\) = \(\dfrac{28}{84}\)

U = \(\dfrac{4}{11}\) = \(\dfrac{4.5}{11.5}\) = \(\dfrac{20}{55}\);   O = \(\dfrac{5}{25}\) = \(\dfrac{5.3}{25.3}\) = \(\dfrac{15}{75}\)

H = \(\dfrac{1}{5}\) = \(\dfrac{1.11}{5.11}\) = \(\dfrac{11}{15}\);    A = \(\dfrac{5}{8}\) = \(\dfrac{5.5}{8.5}\) = \(\dfrac{25}{40}\)

G =  \(\dfrac{-3}{17}\) = \(\dfrac{-3.5}{17.5}\) = \(\dfrac{-15}{85}\); D = \(\dfrac{4}{16}\) = \(\dfrac{4.5}{16.5}\) = \(\dfrac{20}{80}\)

Vì \(\left|x-y\right|\ge0;\left|y+\frac{9}{25}\right|\ge0\Rightarrow\left|x-y\right|+\left|y+\frac{9}{25}\right|\ge0\)

Mà \(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)

\(\Rightarrow\left|x-y\right|=0;\left|y+\frac{9}{25}\right|=0\)

\(\left|y+\frac{9}{25}\right|=0\Rightarrow y=\frac{-9}{25}\)

\(\Rightarrow\left|x-y\right|=\left|x-\frac{-9}{25}\right|=0\Rightarrow x=\frac{-9}{25}\)

=> x = y

mà y + 9/25 = 0

=> y = -9/25

=> x = y = -9/25

\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.24}+...+\frac{4}{253.258}\)

\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)

\(=\frac{4}{5}\cdot\left(\frac{1}{8}-\frac{1}{258}\right)\)

\(=\frac{4}{5}\cdot\frac{125}{1032}\)

\(=\frac{25}{258}\)

\(\frac{4}{8.13}+\frac{4}{13.18}+\frac{4}{18.23}+...+\frac{4}{253.258}\)

\(=\frac{4}{5}\left(\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+...+\frac{5}{253.258}\right)\)

\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+...+\frac{1}{253}-\frac{1}{258}\right)\)

\(=\frac{4}{5}\left(\frac{1}{8}-\frac{1}{258}\right)\)

\(=\frac{4}{5}.\frac{125}{1032}=\frac{25}{258}\)

Từ đề bài, ta suy ra:

\(\frac{2020\left(1+2021\right)}{2019\left(2019+3\right)}=\frac{2020\cdot2022}{2019\cdot2022}=\frac{2020}{2019}\)

\(\frac{2020+2020.2021}{2019^2+2019.3}=\frac{2020.\left(1+2021\right)}{2019.\left(2019+3\right)}=\frac{2020.2022}{2019.2022}=\frac{2020}{2019}=1\frac{1}{2019}\)

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