\(\dfrac{x}{9}\cdot\dfrac{-7}{8}+\dfrac{x}{9}:\dfrac{-8}{17}=\dfrac{5}{7}\)

\(\dfrac{x}{9}\cdot\dfrac{-7}{8}+\dfrac{x}{9}\cdot\dfrac{17}{-8}=\dfrac{5}{7}\)

\(\dfrac{x}{9}\left(\dfrac{-7}{8}+\dfrac{17}{-8}\right)=\dfrac{5}{7}\)

\(\dfrac{x}{9}\cdot\left(-3\right)=\dfrac{5}{7}\)

\(\dfrac{x}{9}=-\dfrac{5}{21}\)

\(x=\dfrac{5\cdot9}{-21}\)

\(x=\dfrac{-15}{7}\)

a)

Điều kiện xác định: \(2n\ne4\Rightarrow n\ne2\)

Để A là phân số thì \(2n\in Z\Rightarrow n\in Z\)

Vậy mọi \(n\in Z,n\ne2\) thì A là phân số.

b)

\(A=\dfrac{2n+2}{2n-4}\)

\(A=\dfrac{2n-4+6}{2n-4}\)

\(A=1+\dfrac{6}{2n-4}\)

\(\Rightarrow2n-4\inƯ\left(6\right)\)

\(Ư\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

Ta loại các ước số lẻ.

Vậy \(n\in\left\{1;-1;3;5\right\}\)

\(\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\)

\(=\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+\left(1-\dfrac{1}{20}\right)+\left(1-\dfrac{1}{30}\right)+\left(1-\dfrac{1}{42}\right)+\left(1-\dfrac{1}{56}\right)+\left(1-\dfrac{1}{72}\right)+\left(1-\dfrac{1}{90}\right)\)

\(=8-\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=8-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)

\(=8-\left(\dfrac{3-2}{2\cdot3}+\dfrac{4-3}{3\cdot4}+\dfrac{5-4}{4\cdot5}+\dfrac{6-5}{5\cdot6}+\dfrac{7-6}{6\cdot7}+\dfrac{8-7}{7\cdot8}+\dfrac{9-8}{8\cdot9}+\dfrac{10-9}{9\cdot10}\right)\)

\(=8-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=8-\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)

\(=8-\dfrac{4}{10}\)

\(=\dfrac{80}{10}-\dfrac{4}{10}=\dfrac{76}{10}=\dfrac{38}{5}\)

\(A=\dfrac{3}{2^2}+\dfrac{8}{3^2}+...+\dfrac{2023^2-1}{2023^2}\)

\(A=\dfrac{2^2-1}{2^2}+\dfrac{3^2-1}{3^2}+...+\dfrac{2023^2-1}{2023^2}\)

\(A=\left(\dfrac{2^2}{2^2}-\dfrac{1}{2^2}\right)+\left(\dfrac{3^2}{3^2}-\dfrac{1}{3^2}\right)+...+\left(\dfrac{2023^2}{2023^2}-\dfrac{1}{2023^2}\right)\)

\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+...+1-\dfrac{1}{2023^2}\)

\(A=2022-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}\right)\)

Mà:

\(\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot3}+...+\dfrac{1}{2023\cdot2023}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2022\cdot2023}\)

Hay:

\(\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot3}+...+\dfrac{1}{2023\cdot2323}< 1-\dfrac{1}{2023}< 1\)

Nên:

\(\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot3}+...+\dfrac{1}{2023\cdot2323}< 1\)

Vậy A không phải là số tự nhiên.

\(\dfrac{12}{-15}=\dfrac{12:-3}{-15:-3}=\dfrac{-4}{5}\)

Ta có: 

Mẫu số chung 2 phân số: 95

\(\dfrac{-4}{5}=\dfrac{-4\cdot19}{5\cdot19}=\dfrac{-76}{95}\)

\(\dfrac{-15}{19}=\dfrac{-15\cdot5}{19\cdot5}=\dfrac{-75}{95}\)

Vì \(-76< -75\) nên\(\dfrac{-76}{95}< \dfrac{-75}{95}\)

Vậy \(\dfrac{12}{-15}< \dfrac{-15}{19}\)

Rút gọn:

\(\dfrac{12}{-15}=\dfrac{12:3}{-15:3}=\dfrac{4}{-5}\)

Ta có:

\(\dfrac{4}{-5}\)  và \(\dfrac{15}{-19}\) (Đổi \(\dfrac{-15}{19}=\dfrac{15}{-19}\))

Quy đồng 2 phân số:

Mẫu số chung: \(95\).

Ta có:

\(\dfrac{4}{-5}=\dfrac{4\cdot\left(-19\right)}{-5\cdot\left(-19\right)}=\dfrac{-76}{95};\dfrac{15}{-19}=\dfrac{15\cdot\left(-5\right)}{-19\cdot\left(-5\right)}=\dfrac{-75}{95}\)

Mà \(\dfrac{76}{95}>\dfrac{75}{95}\Rightarrow\dfrac{-76}{95}< \dfrac{-75}{95}\)

Vậy \(\dfrac{12}{-15}< \dfrac{-15}{19}\)

\(\dfrac{2}{5}\) của 30 là \(30\cdot\dfrac{2}{5}=12\)

\(\dfrac{2}{5}\) của \(30\) là:

\(30\times\dfrac{2}{5}=12\)

Đáp số: \(12\)

\(BC\left(5,6,8\right)=\left\{120;240;360;480;600;...\right\}\)

Trong các số trên, chỉ có số 480 thỏa mãn \(400\le480\le500\)

Vậy số người tham dự buổi tập đồng diễn là: \(480+1=481\) người

Gọi số người dự buổi tập đồng diễn thể dục đó là \(x\left(đk:người,x\inℕ^∗\right)\)

\(x-1⋮5\)

\(x-1⋮6\)

\(x-1⋮8\)

\(400< x-1< 500\)

\(\Rightarrow x-1\in BC\left(5,6,8\right)\)

Ta có:

\(5=5\)

\(6=2\cdot3\)

\(8=2^3\)

\(\Rightarrow BCNN\left(5,6,8\right)=5\cdot3\cdot2^3=120\)

\(\Rightarrow BC\left(5,6,8\right)\in\left\{0;120;240;360;480;600;720;...\right\}\)

\(\Rightarrow x-1\in\left\{0;120;240;360;480;600;720;...\right\}\)

\(\Rightarrow x\in\left\{1;121;241;361;481;601;721;...\right\}\)

Mà \(400< x< 500\Rightarrow x=481\)

Vậy số người chính xác dự buổi tập đồng diễn thể dục là \(481\) người.

\(x\cdot\left(x+7\right)=0\)

Trường hợp 1:

\(x=0\)

Trường hợp 2:

\(x+7=0\)

\(\Rightarrow x=-7\)

Vậy: \(x\in\left\{0;-7\right\}\)

\(x\times\left(x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=0-7=-7\end{matrix}\right.\)

Vậy...

\(x:\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)=100\)

\(\Rightarrow x:\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{99}-\dfrac{1}{100}\right)=100\)

\(\Rightarrow x:\left(1-\dfrac{1}{100}\right)=100\)

\(\Rightarrow x:\dfrac{99}{100}=100\)

\(\Rightarrow x=100\cdot\dfrac{99}{100}\)

\(\Rightarrow x=99\)

\(x:\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)=100\)

\(x:\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=100\)

\(x:\left(\dfrac{1}{1}-\dfrac{1}{100}\right)=100\)

\(x:\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=100\)

\(x:\dfrac{99}{100}=100\)

\(x=100\cdot\dfrac{99}{100}\)

\(x=99\)

\(-45:5\cdot\left(-3-2x\right)=3\)

\(\Rightarrow-9\cdot\left(-3-2x\right)=3\)

\(\Rightarrow-3-2x=-9:3\)

\(\Rightarrow-3-2x=-3\)

\(\Rightarrow-2x=-3+3\)

\(\Rightarrow-2x=0\)

\(\Rightarrow x=0:\left(-2\right)\)

\(\Rightarrow x=0\)