\(S=1+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+...+\dfrac{649}{650}\)

\(=1+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+...+1-\dfrac{1}{650}\)

\(=\left(1+1+...+1\right)-\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{650}\right)\)

\(=\left(1+1+...+1\right)-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{25\cdot26}\right)\)

\(=25-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{25}-\dfrac{1}{26}\right)\)

\(=25-\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=25-\dfrac{12}{26}=25-\dfrac{6}{13}=\dfrac{319}{13}\)

Sửa đề: \(1+\dfrac{1}{8}+\dfrac{1}{24}+...+\dfrac{4}{x\left(x+4\right)}=\dfrac{99}{100}\)

=>\(\dfrac{4}{4}+\dfrac{4}{32}+\dfrac{4}{96}+...+\dfrac{4}{x\left(x+4\right)}==\dfrac{99}{100}\)

=>\(\dfrac{4}{4}+\dfrac{4}{4\cdot8}+\dfrac{4}{8\cdot12}+...+\dfrac{4}{x\left(x+4\right)}=\dfrac{99}{100}\)
=>\(\dfrac{4}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{12}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{99}{100}\)

=>\(\dfrac{5}{4}-\dfrac{1}{x+4}=\dfrac{99}{100}\)

=>\(\dfrac{1}{x+4}=\dfrac{125-99}{100}=\dfrac{26}{100}=\dfrac{13}{50}\)

=>\(x+4=\dfrac{50}{13}\)

=>\(x=-\dfrac{2}{13}\)

Lời giải:
$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}$

$=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$

$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50})$

$=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$

$\Rightarrow (\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}):(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100})=1$

Lời giải:
$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}$

$=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$
$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100})$

$=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{100})-(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50})$

$=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$

$\Rightarrow (\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}):(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100})=1$

\(\dfrac{2a-1}{4}-\dfrac{3a+2}{3}+2a\)

\(=\dfrac{3\left(2a-1\right)-4\left(3a+2\right)+24a}{12}\)

\(=\dfrac{6a-3-12a-8+24a}{12}=\dfrac{18a-11}{12}\)

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.

$=\frac{1}{4}.\frac{-152}{11}+\frac{1}{4}.\frac{-68}{11}$

$=\frac{1}{4}(\frac{-152}{11}+\frac{-68}{11})$

$=\frac{1}{4}.(-20)=-5$

\(A=\dfrac{2n+2}{n+2}+\dfrac{5n+17}{n+2}-\dfrac{3n}{n+2}\)

\(=\dfrac{2n+2+5n+17-3n}{n+2}=\dfrac{4n+19}{n+2}\)

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 $11\vdots n+2$

$\Rightarrow n+2\in\left\{1; -1; 11; -11\right\}$

$\Rightarrow n\in\left\{-1; -3; 9; -13\right\}$

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.

a: Số học sinh khá là \(45\cdot40\%=18\left(bạn\right)\)

Số học sinh còn lại là 45-18=27(bạn)

Số học sinh trung bình là \(27\left(1-\dfrac{5}{9}\right)=27\cdot\dfrac{4}{9}=12\left(bạn\right)\)

b: Số học sinh nữ là \(12:\dfrac{5}{6}=12\cdot\dfrac{6}{5}=\dfrac{72}{5}=14,4\left(bạn\right)\)

=>Đề sai rồi bạn