Tính tổng S=(1+1/2).(1+1/3).(1+1/4).(1+1/5).....(1+1/100)
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\(S=1+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+...+\dfrac{649}{650}\)
\(=1+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+...+1-\dfrac{1}{650}\)
\(=\left(1+1+...+1\right)-\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{650}\right)\)
\(=\left(1+1+...+1\right)-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{25\cdot26}\right)\)
\(=25-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{25}-\dfrac{1}{26}\right)\)
\(=25-\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=25-\dfrac{12}{26}=25-\dfrac{6}{13}=\dfrac{319}{13}\)
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Sửa đề: \(1+\dfrac{1}{8}+\dfrac{1}{24}+...+\dfrac{4}{x\left(x+4\right)}=\dfrac{99}{100}\)
=>\(\dfrac{4}{4}+\dfrac{4}{32}+\dfrac{4}{96}+...+\dfrac{4}{x\left(x+4\right)}==\dfrac{99}{100}\)
=>\(\dfrac{4}{4}+\dfrac{4}{4\cdot8}+\dfrac{4}{8\cdot12}+...+\dfrac{4}{x\left(x+4\right)}=\dfrac{99}{100}\)
=>\(\dfrac{4}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{99}{100}\)
=>\(\dfrac{5}{4}-\dfrac{1}{x+4}=\dfrac{99}{100}\)
=>\(\dfrac{1}{x+4}=\dfrac{125-99}{100}=\dfrac{26}{100}=\dfrac{13}{50}\)
=>\(x+4=\dfrac{50}{13}\)
=>\(x=-\dfrac{2}{13}\)
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Lời giải:
$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$
$=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$
$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}$
$=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50})$
$=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$
$\Rightarrow (\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}):(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100})=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$
$=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$
$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}$
$=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50})$
$=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$
$\Rightarrow (\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}):(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100})=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2a-1}{4}-\dfrac{3a+2}{3}+2a\)
\(=\dfrac{3\left(2a-1\right)-4\left(3a+2\right)+24a}{12}\)
\(=\dfrac{6a-3-12a-8+24a}{12}=\dfrac{18a-11}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(\dfrac{6}{13}\cdot\dfrac{5}{18}+\dfrac{-7}{24}+\dfrac{5}{18}\cdot\dfrac{7}{13}+\dfrac{-5}{24}\)
\(=\dfrac{5}{18}\left(\dfrac{6}{13}+\dfrac{7}{13}\right)+\left(-\dfrac{7}{24}+\dfrac{-5}{24}\right)\)
\(=\dfrac{5}{18}+\dfrac{-12}{24}=\dfrac{5}{18}-\dfrac{1}{2}=\dfrac{-4}{18}=-\dfrac{2}{9}\)
b: \(4\dfrac{1}{20}\left(-\dfrac{2}{3}\right)^2+\left(0,8-\dfrac{8}{15}\right):\dfrac{-4}{7}\)
\(=\dfrac{81}{20}\cdot\dfrac{4}{9}+\left(\dfrac{4}{5}-\dfrac{8}{15}\right)\cdot\dfrac{-7}{4}\)
\(=\dfrac{9}{5}+\dfrac{4}{15}\cdot\dfrac{-7}{4}=\dfrac{9}{5}-\dfrac{7}{15}=\dfrac{20}{15}=\dfrac{4}{3}\)
c: \(\left(3\dfrac{3}{29}-3\dfrac{1}{5}+2\dfrac{7}{11}\right)-\left(2\dfrac{3}{29}-3\dfrac{4}{11}\right)\)
\(=3+\dfrac{3}{29}-3-\dfrac{1}{5}+2+\dfrac{7}{11}-2-\dfrac{3}{29}+3+\dfrac{4}{11}\)
\(=\left(3-3+2-2+3\right)+\left(\dfrac{3}{29}-\dfrac{3}{29}\right)+\left(\dfrac{7}{11}+\dfrac{4}{11}\right)-\dfrac{1}{5}\)
\(=3+1-\dfrac{1}{5}=4-\dfrac{1}{5}=3,8\)
d: \(\dfrac{-1}{4}\cdot\dfrac{152}{11}+\dfrac{1}{4}\cdot\dfrac{-68}{11}\)
\(=\dfrac{1}{4}\left(-\dfrac{152}{11}-\dfrac{68}{11}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{-220}{11}=\dfrac{1}{4}\cdot\left(-20\right)=-5\)
Bài 2:
a: \(\dfrac{5}{6}+\left(5x+\dfrac{3}{2}\right):\dfrac{8}{15}=2\dfrac{1}{12}\)
=>\(\left(5x+\dfrac{3}{2}\right):\dfrac{8}{15}=\dfrac{25}{12}-\dfrac{5}{6}=\dfrac{25-10}{12}=\dfrac{15}{12}=\dfrac{5}{4}\)
=>\(5x+\dfrac{3}{2}=\dfrac{8}{15}\cdot\dfrac{5}{4}=\dfrac{2}{3}\)
=>\(5x=\dfrac{2}{3}-\dfrac{3}{2}=\dfrac{-5}{6}\)
=>\(x=-\dfrac{1}{6}\)
b: \(\dfrac{2}{3}\left(x+\dfrac{9}{5}\right)-\dfrac{3}{10}\left(5x-\dfrac{1}{3}\right)=\dfrac{7}{15}\)
=>\(\dfrac{2}{3}x+\dfrac{6}{5}-\dfrac{3}{2}x+\dfrac{1}{10}=\dfrac{7}{15}\)
=>\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{7}{15}-\dfrac{6}{5}-\dfrac{1}{10}=\dfrac{14-36-3}{30}=\dfrac{-25}{30}=\dfrac{-5}{6}\)
=>\(x\cdot\dfrac{-5}{6}=\dfrac{-5}{6}\)
=>x=1
c: \(\dfrac{3}{5}x-\dfrac{1}{2}=\dfrac{7}{10}\)
=>\(\dfrac{3}{5}x=\dfrac{7}{10}+\dfrac{1}{2}=\dfrac{12}{10}=\dfrac{6}{5}\)
=>x=2
d: \(\dfrac{1}{5}+\dfrac{4}{5}:x=-1\)
=>\(\dfrac{4}{5}:x=-1-\dfrac{1}{5}=\dfrac{-6}{5}\)
=>\(x=-\dfrac{4}{5}:\dfrac{6}{5}=\dfrac{-2}{3}\)
Bài 1:
a.
$=(\frac{6}{13}.\frac{5}{18}+\frac{5}{18}.\frac{7}{13})-(\frac{7}{24}+\frac{5}{24})$
$=\frac{5}{18}(\frac{6}{13}+\frac{7}{13})-\frac{12}{24}$
$=\frac{5}{18}.1-\frac{1}{2}=\frac{5}{18}-\frac{1}{2}=\frac{-2}{9}$
b.
$=\frac{81}{20}.\frac{4}{9}+\frac{4}{15}.\frac{-7}{4}$
$=\frac{9}{5}+\frac{-7}{15}=\frac{4}{3}$
c.
$=3\frac{3}{29}-3\frac{1}{5}+2\frac{7}{11}-2\frac{3}{29}+3\frac{4}{11}$
$=(3-3+2-2+3)+(\frac{3}{29}-\frac{3}{29})+(\frac{7}{11}+\frac{4}{11})$
$=3+0+\frac{11}{11}=3+1=4$
d.
$=\frac{1}{4}.\frac{-152}{11}+\frac{1}{4}.\frac{-68}{11}$
$=\frac{1}{4}(\frac{-152}{11}+\frac{-68}{11})$
$=\frac{1}{4}.(-20)=-5$
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\(A=\dfrac{2n+2}{n+2}+\dfrac{5n+17}{n+2}-\dfrac{3n}{n+2}\)
\(=\dfrac{2n+2+5n+17-3n}{n+2}=\dfrac{4n+19}{n+2}\)
Để A là số nguyên thì \(4n+19⋮n+2\)
=>\(4n+8+11⋮n+2\)
=>\(11⋮n+2\)
=>\(n+2\in\left\{1;-1;11;-11\right\}\)
=>\(n\in\left\{-1;-3;9;-13\right\}\)
Lời giải:
\(A=\frac{2n+2}{n+2}+\frac{5n+17}{n+2}-\frac{3n}{n+2}=\frac{2n+2+5n+17-3n}{n+2}\\
=\frac{4n+19}{n+2}=\frac{4(n+2)+11}{n+2}=4+\frac{11}{n+2}\)
Với $n$ nguyên, để $A$ nguyên thì $11\vdots n+2$
$\Rightarrow n+2\in\left\{1; -1; 11; -11\right\}$
$\Rightarrow n\in\left\{-1; -3; 9; -13\right\}$
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Lời giải:
$4\equiv 1\pmod 3$
$\Rightarrow 4^{2024}+1\equiv 1^{2024}+1\equiv 2\pmod 3$
Một scp khi chia cho 3 thì chỉ có thể có số dư là $0$ hoặc $1$
$\Rightarrow 4^{2024}+1$ không phải số chính phương.
\(S=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=\dfrac{101}{2}\)