a: ĐKXĐ: x<>-3

\(A=\dfrac{4-5x}{x+3}=\dfrac{-5x-15+19}{x+3}=--5+\dfrac{19}{x+3}\)

Để A là số nguyên mà x nhỏ nhất thì x+3=-19

=>x=-22

Còn câu b ạ

Mai mik thi r ạ

Ai đó giúp mik vs nnha

bài 1:

a: \(2n+3⋮n-1\)

=>\(2n-2+5⋮n-1\)

=>\(5⋮n-1\)

=>\(n-1\in\left\{1;-1;5;-5\right\}\)

=>\(n\in\left\{2;0;6;-4\right\}\)

b: \(n^2-2n+4⋮n+1\)

=>\(n^2+n-3n-3+7⋮n+1\)

=>\(7⋮n+1\)

=>\(n+1\in\left\{1;-1;7;-7\right\}\)

=>\(n\in\left\{0;-2;6;-8\right\}\)

c: \(2n^2+n+3⋮2n+1\)

=>\(n\left(2n+1\right)+3⋮2n+1\)

=>\(3⋮2n+1\)

=>\(2n+1\in\left\{1;-1;3;-3\right\}\)

=>\(n\in\left\{0;-1;1;-2\right\}\)
d: \(2n^2-n+2⋮n+2\)

=>\(2n^2+4n-5n-10+12⋮n+2\)

=>\(12⋮n+2\)

=>\(n+2\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(n\in\left\{-1;-3;0;-4;1;-5;2;-6;4;-8;10;-14\right\}\)

a)(n²-3n+1)⋮(n-2)
Vì (n-2)⋮(n-2)
⇒n.(n-2)⋮(n-2)
⇒[(n²-3n+1)-n.(n-2)]⋮(n-2)
⇒[(n²-3n+1)-(n²-2n)]⋮(n-2)
⇒[n²-3n+1-n²+2n⁰ ]⋮(n-2)
⇒(-n+1):(n-2)
⇒-(n-1)⋮(n-2)
⇒(n-2+1)⋮(n-2)
 Vì (n-2)⋮(n-2)
⇒1⋮(n-2)
 Vì n nguyên
⇒(n-2)ϵƯ(1)={-1;1}Ư

1: \(\dfrac{15}{7}\cdot\dfrac{-3}{21}=\dfrac{15}{21}\cdot\dfrac{-3}{7}=\dfrac{-3}{7}\cdot\dfrac{5}{7}=-\dfrac{15}{49}\)

2: \(\dfrac{34}{-11}\cdot\left(\dfrac{1}{3}-\dfrac{5}{6}\right)^2+\dfrac{10}{11}\cdot\left(0,125\right)\)

\(=\dfrac{-34}{11}\left(\dfrac{2}{6}-\dfrac{5}{6}\right)^2+\dfrac{10}{11}\cdot\dfrac{1}{8}\)

\(=\dfrac{-34}{11}\cdot\dfrac{1}{4}+\dfrac{10}{88}\)

\(=\dfrac{-34}{44}+\dfrac{10}{88}=\dfrac{-34}{44}+\dfrac{5}{44}=\dfrac{-29}{44}\)

3: \(\left(\dfrac{3}{5}-\dfrac{1}{15}\right)\cdot\dfrac{-8}{7}-\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{4}\right)^2\)

\(=\left(\dfrac{9}{15}-\dfrac{1}{15}\right)\cdot\dfrac{-8}{7}-\dfrac{1}{3}\left(\dfrac{1}{4}\right)^2\)

\(=\dfrac{8}{15}\cdot\dfrac{-8}{7}-\dfrac{1}{3}\cdot\dfrac{1}{16}\)

\(=-\dfrac{64}{105}-\dfrac{1}{48}=-\dfrac{353}{560}\)

\(-\dfrac{3}{11}:\dfrac{2}{9}+\dfrac{-3}{11}\cdot\dfrac{7}{3}\)

\(=-\dfrac{3}{11}\cdot\dfrac{9}{2}+\dfrac{-3}{11}\cdot\dfrac{7}{3}\)

\(=-\dfrac{3}{11}\left(\dfrac{9}{2}+\dfrac{7}{3}\right)\)

\(=-\dfrac{3}{11}\cdot\dfrac{27+14}{6}=\dfrac{-3}{6}\cdot\dfrac{41}{11}=\dfrac{-1}{2}\cdot\dfrac{41}{11}=\dfrac{-41}{22}\)

a: Số tiền lãi bác Long nhận được sau 1 năm là:

\(120000000\cdot6,5\%=7800000\left(đồng\right)\)

Tổng số tiền nhận được là:

120000000+7800000=127800000(đồng)

b: Lãi suất gửi không thời hạn là:

6,5%-2%=4,5%

Số tiền cả gốc lẫn lãi bác Long nhận được là:

\(120000000\left(1+4,5\%\right)=125400000\left(đồng\right)\)

\(M=1+\dfrac{6}{2\cdot5}+\dfrac{10}{5\cdot10}+\dfrac{14}{10\cdot17}+\dfrac{18}{17\cdot26}\)

\(=1+2\left(\dfrac{3}{2\cdot5}+\dfrac{5}{5\cdot10}+\dfrac{7}{10\cdot17}+\dfrac{9}{17\cdot26}\right)\)

\(=1+2\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{26}\right)\)

\(=1+2\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=1+2\cdot\dfrac{12}{26}=1+\dfrac{24}{26}=\dfrac{50}{26}=\dfrac{25}{13}\)

Sửa đề: \(\dfrac{1}{x}\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}\)

=>\(\dfrac{1}{x}\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)

=>\(\dfrac{1}{x}\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)

=>\(\dfrac{1}{x}\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)

=>\(\dfrac{1}{x}=100\)

=>x=1/100

Gọi \(d=ƯC\left(2n+1;4n-2\right)\)

Do \(2n+1\) lẻ \(\Rightarrow d\) lẻ

Ta có: \(\left\{{}\begin{matrix}2n+1⋮d\\4n-2⋮d\end{matrix}\right.\)

\(\Rightarrow2\left(2n+1\right)-\left(4n-2\right)⋮d\)

\(\Rightarrow4⋮d\Rightarrow\left[{}\begin{matrix}d=1\\d=2\\d=4\end{matrix}\right.\)

Mà d lẻ \(\Rightarrow d=1\)

\(\Rightarrow\dfrac{2n+1}{4n-2}\) tối giản

d; \(\dfrac{2x-1}{12}\) = \(\dfrac{5}{3}\)

    2\(x\) - 1 = \(\dfrac{5}{3}\).12

   2\(x\) - 1 = 20

  2\(x\) = 20 + 1

  2\(x\) = 21

    \(x\) = 21 : 2

    \(x=\dfrac{21}{2}\)

Vậy \(x=\dfrac{21}{2}\)

e; \(\dfrac{x}{3}\) - \(\dfrac{1}{4}\) = \(\dfrac{-5}{6}\)

    \(\dfrac{x}{3}\)        =  \(\dfrac{-5}{6}\) + \(\dfrac{1}{4}\)

    \(\dfrac{x}{3}\)        = - \(\dfrac{7}{12}\)

    \(x\)        = - \(\dfrac{7}{12}\) x 3

     \(x\)     = - \(\dfrac{7}{4}\)

Vậy  \(x\) = - \(\dfrac{7}{4}\)

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