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a: \(x\in B\left(5\right)\)
=>\(x\in\left\{0;5;10;15;20;25;30;35;40;...\right\}\)
mà 20<=x<=36
nên \(x\in\left\{20;25;30;35\right\}\)
b: \(x\inƯ\left(20\right)\)
=>\(x\in\left\{1;2;4;5;10;20\right\}\)
mà x>8
nên \(x\in\left\{10;20\right\}\)
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =-\dfrac{12}{5}:\dfrac{7}{5}=\dfrac{-12}{7}\)
=>-10<=x<=-12/7
hay \(x\in\left\{-10;-9;-8;-7;-6;-5;-4;-3;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{1}{8}\)
=>-13/9<=x<=-1/12
hay \(x=-1\)
`-1/5<=x/8<=1/4`
`=>8* -1/5<=x<=1/4*8`
`=>-8/5<=x<=2`
Mà `x in ZZ`
`=>x in {-1,0,1,2}`
−1/5≤x8≤1/4-15≤x8≤14
⇒8⋅−1/5≤x≤14⋅8⇒8⋅-15≤x≤14⋅8
⇒−85≤x≤2⇒-85≤x≤2
Mà x∈Zx∈ℤ
⇒x∈{−1,0,1,2}
a) Để \(x\le6\left(x\in N\right)\) thì \(x=0,1,2,3,4,5,6\)
b) Để \(35\le x\le39\) thì \(x=35,36,37,38,39\)
c) Để \(216< x\le219\) thì \(x=217,218,219\)
Bài 2:
a) Để 3369 < 33*9 < 3389 thì * = 7
b) Để 2020 \(\le\) 20*0 < 2040 thì x = 2, 3
\(#Wendy.Dang\)
\(a) \) \(-5\le x+8\le5\)
\(\Leftrightarrow\)\(x+8\in\left\{\pm5;\pm4;\pm3;\pm2;\pm1;0\right\}\)
\(x\in\left\{-3;-13;-4;-12;-5;-11;-6;-10;-7;-9;-8\right\}\)
\(b) \)
\(2004\le\left|2x\right|\le2010\)
\(\Leftrightarrow\left|2x\right|\in\left\{2004;2006;2008;2010\right\}\)
\(\Leftrightarrow2x\in\left\{\pm2004;\pm2006;\pm2008;\pm2010\right\}\)
\(\Leftrightarrow x\in\left\{\pm1002;\pm1003;\pm1004;\pm1005\right\}\)
Giải:
Theo bài ra ta có:
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{12}\)
\(\Rightarrow-3\le x\le\frac{23}{12}\)
\(\Rightarrow x\varepsilon\left\{-2;-1;0;1\right\}\)
\(\frac{-5}{6}+\frac{16}{6}+-\frac{29}{6}\le x\le\frac{-6}{12}+\frac{24}{12}+\frac{5}{12}\)
=>-3\(\le\) x\(\le\) 23/12
=> x thuộc{-2-1;0;1}
\(-10\le\left|5-x\right|\le2\)
\(\Rightarrow\left|5-x\right|=1;2\)
\(\Rightarrow5-x=\pm1\) hoặc \(5-x=\pm2\)
+) \(5-x=1\Rightarrow x=4\)
+) \(5-x=-1\Rightarrow x=6\)
+) \(5-x=2\Rightarrow x=3\)
+) \(5-x=-2\Rightarrow x=7\)
Vậy \(x\in\left\{4;6;3;7\right\}\)
\(-10\le\left|5-x\right|\le2\\ \Rightarrow\left|5-x\right|=1;2\\ \Rightarrow5-x=\pm1ho\text{ặ}c5-x=\pm2\\ \)
+) \(5-x=1\\ \Rightarrow x=4\\ \)
+) \(5-x=-1\Rightarrow x=6\)
+) \(5-x=2\Rightarrow x=3\)
+) \(5-x=-2\Rightarrow x=7\\ \)
Vậy \(x\in\left\{4;6;3;7\right\}\)
a)x \(\in\){ -5; -4; -3; -2; -1;..;2005;2006;2007;2008;2009;2010}
b)x\(\in\){2004;2005;2006;2007;2008;2009;2010}
c)IxI\(\in\){0;1;2;3;4;5;6}
\(\Rightarrow\)x\(\in\){-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6}
-5 \(\le\)x\(\le\)2010 nên x \(\in\){-5; -4; -3; -2;...........; 2007; 2008; 2009; 2010}
2004\(\le\)x\(\le\)2010 nên x \(\in\){2004; 2005; 2006; 2007; 2008; 2009; 2010}
IxI \(\le\)6 nên x\(\in\){-6; -5; -4; -3; -2; -1; 0; 1; 2; 3; 4; 5; 6}