Tìm xϵN biết:
(n+5) ⋮(n+2)
(^-^)
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\(3n+14⋮n+2\)
=>\(3n+6+8⋮n+2\)
=>\(8⋮n+2\)
=>\(n+2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
=>\(n\in\left\{-1;-3;0;-4;2;-6;6;-10\right\}\)
mà n>=0
nên \(n\in\left\{0;2;6\right\}\)
a, 2\(^x\) - 15 = 17
2\(^x\) = 17 + 15
2\(^x\) = 32
2\(^x\) = 25
\(x\) = 5
b, (2\(x\) - 11)5 = 24.32 + 99
(2\(x\) - 11)5 = 16.9 + 99
(2\(x\) - 11)5 = 144 + 99
(2\(x\) - 11)5 = 243
(2\(x\) - 11)5 = 35
2\(x\) - 11 = 3
2\(x\) = 3 + 11
2\(x\) = 14
\(x\) = 14: 2
\(x\) = 7
c, \(x^{10}\) = 1\(^x\)
\(x^{10}\) = 1
\(x^{10}\) = 110
\(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(x\) \(\in\) { -1; 1}
A) \(...\Rightarrow2^x=32=2^5\Rightarrow x=5\)
B) \(...\Rightarrow\left(2x-11\right)^5=243=3^5\)
\(\Rightarrow2x-11=5\Rightarrow2x=16\Rightarrow x=8\)
C) \(...\Rightarrow x^{10}=1=x^0\Rightarrow x=1\)
\(7⋮x-1\)
\(\Rightarrow x-1\inƯ\left(7\right)\)
Ta có:
\(Ư\left(7\right)=\left\{1;7\right\}\)
\(\Rightarrow x-1\in\left\{1;7\right\}\)
\(\Rightarrow x\in\left\{2;8\right\}\)
\(a,A=\left\{0;1;2;3;4\right\}\\ b,B=\left\{-16;-13;-10;-7;-4;-1;2;5;8\right\}\\ c,C=\left\{-9;-8;-7;...;7;8;9\right\}\\ d,x^2-3x+1=0\\ \Delta=9-4=5\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{5}}{2}\\x=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\\ \Leftrightarrow D=\left\{\dfrac{3-\sqrt{5}}{2};\dfrac{3+\sqrt{5}}{2}\right\}\)
\(e,2x^3-5x^2+2x=0\\ \Leftrightarrow x\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\dfrac{1}{2}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow E=\left\{0;2\right\}\\ f,F=\left\{0;3;6;9;12;15;18\right\}\)
Lời giải:
Mệnh đề phủ định:\(\forall x\in\mathbb{N}: x^2+5x+6\neq 0\)
Mệnh đề đúng vì \(x\in\mathbb{N}\Rightarrow x\geq 0\)
\(\Rightarrow x^2+5x+6\geq 0^2+5.0+6>0\)
Lời giải:
$A=\left\{0; 1;2;3;4\right\}$
$B=\left\{3\right\}$
$C=\left\{1;3\right\}$
$D=\left\{x\in\mathbb{N}|x=2k+1, \forall k\in\mathbb{N}\right\}$
Như vậy:
$B\subset C\subset A$
$B\subset C\subset D$
45,90,150 đều chia hết cho x
\(\Rightarrow x\inƯC\left(45;90;150\right)\\ 45=5.3^2;90=2.5.3^2;150=2.5^2.3\\ \RightarrowƯCLN\left(45;90;150\right)=5.3=15\\ \Rightarrow x\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)
Vì x< 10 => x=1 hoặc x=3 hoặc x=5
\(\frac{3x+8}{x+1}=3+\frac{5}{x+1}\) ĐKXĐ: \(x\ne-1\)
để \(\frac{3x+8}{x+1}\in Z\Leftrightarrow x+1\inƯ_{\left(5\right)}=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x=-2\\x=4\\x=-6\end{matrix}\right.\)
\(\Rightarrow A=\left\{-2;-6;0;4\right\}\)
\(B=\left\{x\in N|\left|x+2\right|< 5\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2< 5\\x+2< -5\end{matrix}\right.\) (x+2<-5 loại vì \(x\in N\) )
\(\Rightarrow B=\left\{0;1;2\right\}\)
\(\Rightarrow A\cup B=\left\{0;1;2;-2;-6;4\right\}\)
\(A\cap B=\left\{0\right\}\)
A\B={-2;-6;4}