a: \(x\in\left\{-6;-5;-4;-3;-2;-1;0;1;2;3\right\}\)

c: \(x\in\left\{-4;-3;-2;-1\right\}\)

\(a,\Leftrightarrow7⋮x-1\Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\\ b,\Leftrightarrow\dfrac{x-1+2}{x-1}\in Z\Leftrightarrow1+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow2⋮x-1\Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{-1;0;2;3\right\}\)

a) để 7/x-1 thuộc Z 

=> (x-1) thuộc ước 7(+-1;+-7)

x-1  -1     1      -7      7

x      0     2       -6     8

a) \(\dfrac{7}{x-1}\in Z\Leftrightarrow x-1\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)

\(\Rightarrow x\in\left\{2;0;8;-6\right\}\)

b) \(\dfrac{x+1}{x-1}\in Z\Rightarrow\dfrac{x-1}{x-1}+\dfrac{2}{x-1}=1+\dfrac{2}{x-1}\in Z\)

\(\Rightarrow x-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

\(\Rightarrow x\in\left\{2;0;3;-1\right\}\)

\(a,\dfrac{x}{5}=-\dfrac{3}{y}\Rightarrow xy=-15\\ \Rightarrow xy=-1\cdot15=-15\cdot1=-5\cdot3=-3\cdot5\\ \Rightarrow\left(x;y\right)=\left\{\left(-1;-15\right);\left(1;-15\right);\left(15;-1\right);\left(-15;1\right);\left(3;-5\right);\left(-5;3\right);\left(5;-3\right);\left(-3;5\right)\right\}\)\(g,-\dfrac{11}{x}=\dfrac{y}{3}\\ \Rightarrow xy=-33\\ \Rightarrow xy=-3\cdot11=-11\cdot3=-1\cdot33=-33\cdot1\\ \Rightarrow\left(x;y\right)=\left\{\left(-3;11\right);\left(11;-3\right);\left(-11;3\right);\left(3;-11\right);\left(-1;33\right);\left(33;-1\right);\left(-33;1\right);\left(1;-33\right)\right\}\)

a: \(\left(x,y\right)\in\left\{\left(1;2\right);\left(-1;-2\right);\left(2;1\right);\left(-2;-1\right)\right\}\)

\(x^{2022}\ge0\Leftrightarrow A=2021-x^{2022}\le2021\\ A_{max}=2021\Leftrightarrow x=0\)

a: a=0

b: a=48

b: a=0

a: x+2017=-1

hay x=-2018

\(6,\)

\(a,x+2017=-1\)

\(\Rightarrow x=-2018\)

Vậy: \(x=-2018\)

\(b,y-\left(-100\right)=1\)

\(\Rightarrow y+100=1\)

\(\Rightarrow y=-99\)

Vậy: \(y=-99\)

vì a là số nhỏ nhất :5 dư 3 nên a=8.mà a :7 dư 4 nên a=18.thay vào ta được 18:5=3(dư3),18:7=2(dư4)

1: 

Ta có: \(D=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{53\cdot55}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{53\cdot55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{11}{55}-\dfrac{1}{55}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{2}{11}=\dfrac{3}{11}\)

2) Để A là số nguyên dương thì 

\(\left\{{}\begin{matrix}x+2⋮x-5\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5+7⋮x-5\\x>5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7⋮x-5\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5\inƯ\left(7\right)\\x>5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-5\in\left\{1;-1;7;-7\right\}\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;4;12;-2\right\}\\x>5\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{6;12\right\}\)

Giải:

1)D=3/5.7+3/7.9+3/9.11+...+3/53.55

   D=3/2.(2/5.7+2/7.9+2/9.11+...+2/53.55)

   D=3/2.(1/5-1/7+1/7-1/9+1/9-1/11+...+1/53-1/55)

   D=3/2.(1/5-1/55)

   D=3/2.2/11

   D=3/11

a, x=-1991   b, y=-199   c, x=10    d, x=0