C={2x | x\(\in\)N* và x\(\le\)5}
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a: A={0;1;2;3}
b: B={-16;-13;-10;-7;-4;-1;2;5;8}
c: C={-9;-8;-7;...;7;8;9}
d: \(D=\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\)\(A=\left\{x\in R|x< 3\right\}\Rightarrow A=\left(\text{ -∞;3}\right)\)
\(B=\left\{-1;0;1;2;3;4;5\right\}\)
\(\Rightarrow A\cap B=\left\{-1;0;1;2\right\}\)
\(b,x=-1\Rightarrow y=1-2\left(-1\right)+m=m+3\)
\(x=1\Rightarrow y=1-2+m=m-1\)
\(\Rightarrow C=(m-1;m+3]\subset A\)
\(\Rightarrow C\subset A\Leftrightarrow m+3< 3\Leftrightarrow m< 0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a) \) \(-5\le x+8\le5\)
\(\Leftrightarrow\)\(x+8\in\left\{\pm5;\pm4;\pm3;\pm2;\pm1;0\right\}\)
\(x\in\left\{-3;-13;-4;-12;-5;-11;-6;-10;-7;-9;-8\right\}\)
\(b) \)
\(2004\le\left|2x\right|\le2010\)
\(\Leftrightarrow\left|2x\right|\in\left\{2004;2006;2008;2010\right\}\)
\(\Leftrightarrow2x\in\left\{\pm2004;\pm2006;\pm2008;\pm2010\right\}\)
\(\Leftrightarrow x\in\left\{\pm1002;\pm1003;\pm1004;\pm1005\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=\left\{13;14;15\right\}\)
b, \(B=\left\{1;2;3;4\right\}\)
c, \(C=\left\{13;14;15\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a:
TH1: |x+1|=1 và |y-2|=1
=>5|x+1|=5 và |y-2|=2
=>|x+1|=1 và |y-2|=2
=>\(\left\{{}\begin{matrix}x+1\in\left\{1;-1\right\}\\y-2\in\left\{2;-2\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{4;0\right\}\end{matrix}\right.\)
TH2: |x+1|=0 và |y-2|=7
=>y-2=7 hoặc y-2=-7 và x=-1
=>y=7 hoặc y=-5 và x=-1
b: TH1:
4|2x+5|=4 và |y+3|=1
=>|2x+5|=1 và |y+3|=1
=>\(\left\{{}\begin{matrix}2x+5\in\left\{1;-1\right\}\\y+3\in\left\{1;-1\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{-2;-3\right\}\\y\in\left\{-2;-4\right\}\end{matrix}\right.\)
TH2: |2x+5|=0 và |y+3|=5
=>2x+5=0 và |y+3|=5
=>\(\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y\in\left\{2;-8\right\}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) M = {10; 11; 12; 13; 14}
b) K = {1; 2; 3}
c) L = {0; 1; 2; 3}
a) \(M=\left\{10;11;12;13;14\right\}\)
b) \(K=\left\{1;2;3\right\}\)
c) \(L=\left\{0;1;2;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1+3+5+...+x=1600
=(x+1).[(x-1):2+1] /2 =1600
=(x+1).(x+1) /2 =1600
=(x+1)^2:2=40^2
=(x+1):2=40
=x+1=80
=x=79
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a \(2x+2>4\\ \Leftrightarrow2\left(x+1\right)>4\\ \Leftrightarrow x+1>2\\ \Leftrightarrow x>1\)
b \(3x+2>-5\\ \Leftrightarrow3x>-7\\ \Leftrightarrow x>\dfrac{-7}{3}\)
c \(10-2x>2\\ \Leftrightarrow2\left(5-x\right)>2\\ \Leftrightarrow5-x>1\\ \Leftrightarrow-x>-4\\ \Leftrightarrow x< 4\)
d \(1-2x< 3\\ \Leftrightarrow-2x< 2\\ \Leftrightarrow2x>2\\ \Leftrightarrow x>1\)
a)2x+2>4
<=> 2x>4-2
<=>2x>2
<=>x>1
Vậy...
b)3x+2>-5
<=>3x>-5-2
<=>3x>-7
<=>x>\(\dfrac{-7}{3}\)
Vậy...
c)10-2x>2
<=>-2x>-10+2
<=>-2x>-8
<=>x<4
Vậy...
d)1-2x<3
<=>-2x<3-1
<=>-2x<2
<=>x>-1
Vậy...
e)10x+3-5\(\le\)14x+12
<=>10x-2\(\le\)14x+12
<=>10x-14x\(\le\)2+12
<=>-4x\(\le\)14
<=>x\(\ge\)\(\dfrac{-7}{2}\)
Vậy...
f)(3x-1)<2x+4
<=> 3x-2x<1+4
<=>x<5
Vậy...
x={1;2;3;4;5}