\(\left(x-7\right)\left(x+3\right)< 0\)

\(\orbr{\begin{cases}\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}}\\\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x< 7\\x>-3\end{cases}}\\\hept{\begin{cases}x>7\\x< -3\end{cases}}\end{cases}}\Leftrightarrow-3< x< 7\)

\(3-\frac{x}{5}-x=\left(-\frac{3}{5}\right)^2\)

đề như thế này à

\(\frac{3-x}{5-x}=\left(-\frac{3}{5}\right)^2\)

=> \(\frac{3-x}{5-x}=\frac{9}{25}\)

=> (3 - x). 25 = 9.(5 - x)

=> 75 - 25x = 45 - 9x

=> 75 - 45 = -9x + 25x

=> 30 = 16x

=> x = 30  : 16

=> x = 15/8

\(\frac{2016}{2017}< 1\)

\(\frac{2016}{2017}< 1\)

\(\frac{2017}{2018}< 1\)

\(=>\frac{2017}{2018}+\frac{2016}{2017}< 1\)

\(x\in\left\{0;1;2;3;4;5;6;7;8;9\right\}\)

\(0\le\frac{x}{5}\le2\Rightarrow\frac{x}{5}\in\left\{0;1;2\right\}\)

\(\Rightarrow x\in\left\{0;5;10\right\}\)

\(A=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)

\(A=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)

\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)

\(A=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)

\(A=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)

\(A=\frac{2}{3}.\left(1-\frac{1}{19}\right)\)

\(A=\frac{2}{3}.\frac{18}{19}\)

\(\Rightarrow A=\frac{12}{19}\)

\(A=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{104}+\frac{1}{152}\)

\(A=\frac{2}{4}+\frac{2}{28}+\frac{2}{70}+\frac{2}{130}+\frac{2}{208}+\frac{2}{304}\)

\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)

\(A=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)

\(A=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)

\(A=\frac{2}{3}.\left(1-\frac{1}{19}\right)\)

\(A=\frac{2}{3}.\frac{18}{19}\)

\(A=\frac{12}{19}\)

\(x-\left(24-x\right)=x-19\)

\(x-24+x-x=19\)

\(x+x-x=24+19\)

\(x=43\)

x - ( 24 - x ) = x -19

=> x - 24 + x =  x - 19

=>  -24 = x - 19

=> -x = 24 - 19
=> -x = 5

=> x = -5

\(\frac{1}{2}\left(x-\frac{1}{3}\right)-3\left(x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(x-\frac{1}{3}\right).2-3\left(x-\frac{1}{3}\right).2=0.2\)

\(\Leftrightarrow x-\frac{1}{3}-6\left(x-\frac{1}{3}\right)=0\)

\(\Leftrightarrow-5x+\frac{5}{3}=0\)

\(\Leftrightarrow-5x=0-\frac{5}{3}\)

\(\Leftrightarrow-5x=-\frac{5}{3}\)

\(\Leftrightarrow x=\left(-\frac{5}{3}\right):\left(-5\right)\)

\(\Leftrightarrow x=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}\)

1/2x(x-1/3) là 1/2 nhân (x-1/3) à 

\(\frac{27^3.4^5}{6^8}:\left(\frac{5^5.2^4}{10^4}:\frac{6^4}{2^6.3^4}\right)\)

\(=\frac{3^9.2^{10}}{6^8}:\left(5:\frac{1}{2^2}\right)\)\(=3.2^2:20=\frac{12}{20}=\frac{3}{5}\)

\(\frac{27^3.4^5}{6^8}:\left(\frac{5^5.2^4}{10^4}:\frac{6^4}{2^6.3^4}\right)\)

\(=\frac{\left(3^3\right)^3.\left(2^2\right)^5}{\left(3.2\right)^8}:\left(\frac{5^5.2^4}{\left(5.2\right)^4}:\frac{\left(2.3\right)^4}{2^6.3^4}\right)\)

\(=\frac{3^9.2^{10}}{3^8.2^8}:\left(\frac{5^5.2^4}{5^4.2^4}:\frac{2^4.3^4}{2^6.3^4}\right)\)

\(=3.2^2:\left(5:\frac{1}{2^2}\right)\)

\(=3.4:\left(5.4\right)\)

\(=12:20\)

\(=\frac{12}{20}=\frac{3}{5}\)

Để \(P\in Z\)thì \(n\in Z\)

\(P=\frac{2n+5}{n+3}\)

\(\Rightarrow P=\frac{2n+6-1}{n+3}\)

\(\Rightarrow P=2+\frac{-1}{n+3}\)

Mà \(n\in Z;-1⋮n+3\)

\(\Rightarrow n+3\in\left\{-1;1\right\}\)

\(\Rightarrow n\in\left\{-4;-2\right\}\)

3. Từ đề bài, ta có :

\(\frac{x}{9}-\frac{1}{18}=\frac{3}{y}\)

\(\Rightarrow\frac{2x-1}{18}=\frac{3}{y}\)

\(\Rightarrow\left(2x-1\right).y=18.3=54\)

Mà \(2x-1\)là số lè.

\(\Rightarrow\)Ta có bảng sau :

Vậy ta tìm được 3 cặp số ( x;y ) thỏa mãn đề bài là : ( 1;54 ) ; ( 14;2 ) ; ( 5;6 )

P/s : Bài 2 k làm được thì ib mk nhé -.-