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Lời giải:
a) \(12\vdots x+1\Rightarrow x+1\in \text{Ư}(12)\)
Mà \(x\in\mathbb{N}\Rightarrow x+1\in\mathbb{N}\). Do đó \(x+1\in \left\{1; 2;3;4;6;12\right\}\)
\(\Rightarrow x\in \left\{0; 1;2;3;5;11\right\}\)
b)
\(x+5\vdots x+1\)
\(\Rightarrow (x+1)+4\vdots x+1\)
\(\Rightarrow 4\vdots x+1\Rightarrow x+1\in \text{Ư}(4)\). Mà \(x\in \mathbb{N}\Rightarrow x+1\in \mathbb{N}\)
Do đó: \(x+1\in \left\{1;2;4\right\}\)
\(\Rightarrow x\in \left\{0; 1;3\right\}\)
a) \(\frac{3}{4}+\frac{1}{4}:x=-\frac{2}{5}\)
\(\frac{1}{4}:x=-\frac{23}{20}\)
\(x=-\frac{5}{23}\)
b) \(\frac{2}{5}-\left(\frac{1}{5}+x\right)=0,5\)
\(\frac{1}{5}+x=-\frac{1}{10}\)
\(x=-\frac{3}{10}\)
a) 3/4+1/4:x=2/-5
1/4:x=-2/5-3/4
1/4:x=-23/20
x=1/4:-23/20
x=-5/23
b)2/5-(1/5+x)=0,5
2/5-(1/5+x)=1/2
1/5+x=2/5-1/2
1/5+x=-1/10
x=-1/10-1/5
x=-3/10
dấu / là dấu phân số
Để \(A\) là số nguyên thì \(\left(n+1\right)⋮\left(n-3\right)\)
Ta có :
\(n+1=n-3+4\) chia hết cho \(n-3\) \(\Rightarrow\) \(4⋮\left(n-3\right)\) \(\left(n-3\right)\inƯ\left(4\right)\)
Mà \(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Suy ra :
\(n-3\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(4\) | \(-4\) |
\(n\) | \(4\) | \(2\) | \(5\) | \(1\) | \(7\) | \(-1\) |
Vậy \(n\in\left\{4;2;5;1;7;-1\right\}\)
a, D={1; 2; 3; 6}
b, B={-4; -3; -2; -1; 0; 1; 2; 3; 4}
c, C={-3; -2; -1; 0; 1; 2; 3}
a, \(x\in\left\{1,2,3,4,6,8,12,24\right\}\)
b, \(x\in\left\{-3,-2,-1,0,1,2,3,4\right\}\)
c, \(x\in\left\{-3,-2,-1,0,1,2,3\right\}\)
b) 52-\(|\)x\(|\)=-80
\(|\)x\(|\)=52-(-80)
\(|\)x\(|\)=52+80
\(|\)x\(|\)=132
Vậy x=-132
a) 80 \(⋮\)x
=> x \(\inƯ\left(80\right)=\left\{1;2;4;5;8;10;16;20;40;80\right\}\)
Mà x > 20 nên \(x\notin\left\{1;2;4;5;8;10;16;20\right\}\)
Vậy \(x\in\left\{40;80\right\}\)
b) \(x\inƯ\left(100\right)=\left\{1;2;5;10;20;25;50;100\right\}\)
Mà 5 < x < 20 => \(x\notin\left\{1;2;5;20;25;50;100\right\}\)
Vậy x = 10
c) \(x⋮17\)=> x \(\in\)B(17) = { \(0;17;34;51;...\)}
Mà 10 < x < 30 => \(x\notin\left\{0;34;51;...\right\}\)
=> x = 17
d) \(x\inƯ\left(45\right)\)
=> \(x\in\left\{1;3;5;9;15;45\right\}\)
Mà x > 5 => x \(\notin\left\{1;3;5\right\}\)
Vậy \(x\in\left\{9;15;45\right\}\)
e) \(x\in B\left(15\right)=\left\{0;15;30;45;60;75;90;...;195;210...\right\}\)
Mà \(100\le x\le200\)=> \(x\notin\left\{0;15;30;...;90\right\}\)
Vậy \(x\in\left\{105;120;135;150;165;180;195\right\}\)
Còn câu j tự làm
a) \(x\in B\left(3\right)=\left\{0;3;6;9;12;15;18;21;24;...;63;66;...\right\}\)
Mà 21 \(\le x\le\)65 => \(x\notin\left\{0;3;6;9;12;15;18;66;...\right\}\)
Vậy \(x\in\left\{21;24;...;63\right\}\)
b) \(x⋮17\)
=> x là bội của 17 => x \(\in B\left(17\right)=\left\{0;17;34;51;68;...\right\}\)
Mà \(0\le x\le60\Rightarrow x\in\left\{0;17;34;51\right\}\)
Vậy : ...
c) \(12⋮x\)=> x \(\inƯ\left(12\right)=\left\{1;2;3;4;6;12\right\}\)
d) \(x\inƯ\left(30\right)=\left\{1;2;3;5;6;10;15;30\right\}\)
Mà x \(\ge0\)thì nguyên dàn x đã tìm ở trên :)
e) \(x⋮7\)
=> x là bội của 7 => x \(\in\)B(7) = {0;7;14;21;28;35;42;49;56;...}
Mà x \(\le\)50 thì x \(\in\){0;7;14;21;28;35;42;49}
a) \(12⋮x+1\)
\(\Rightarrow x+1\inƯ\left(12\right)\)
mà \(Ư\left(12\right)=\left\{1;2;3;4;6;12\right\}\)
\(\Rightarrow\hept{\begin{cases}x+1=1;x+1=4\\x+1=2;x+1=6\\x+1=3;x+1=12\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0;x=3\\x=1;x=5\\x=2;x=11\end{cases}}\)
b) \(x+5⋮x+1\)
\(\Rightarrow x+4+1⋮x+1\)
\(\Rightarrow\left(x+1\right)+4⋮x+1\)
\(\Rightarrow4⋮x+1\)
\(\Rightarrow x+1\inƯ\left(4\right)\)
mà \(Ư\left(4\right)=\left\{1;2;4\right\}\)
\(\Rightarrow\hept{\begin{cases}x+1=1\\x+1=2\\x+1=4\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=1\\x=3\end{cases}}\)