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\(c,20=2^2\cdot5\\ 45=3^2\cdot5\\ ƯCLN\left(20,45\right)=5\\ \RightarrowƯC\left(20,45\right)=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\\ C=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(d,\left(6x^2-7x+1\right)\left(x^3-x\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6x-1\right)x\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{6}\\x=1\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow D=\left\{-1;0;\dfrac{1}{6};1\right\}\)
Sửa: \(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{6}\\x=1\\x=-1\end{matrix}\right.\Leftrightarrow D=\left\{-1;0;\dfrac{1}{6};1\right\}\)
\(x^4-6x^2+8=0\\ \Leftrightarrow x^4-2x^2-4x^2+8=0\\ \Leftrightarrow x^2\left(x-2\right)-4\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)^2\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(F=\left\{-2;2\right\}\)
Ta có: \(x^4-6x^2+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)+2x^2\left(x-2\right)-2x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
Vậy \(F=\left\{2;-2;\sqrt{2};-\sqrt{2}\right\}\)
a: x+2<=1
=>x<=-1
=>E={;...;-2;-1}
b: 3<n^2<30
mà n thuộc N
nên \(n^2\in\left\{4;9;16;25\right\}\)
=>\(F=\left\{2;3;4;5\right\}\)
g: -4<x<12
mà x chia hết cho 3(x=3k; k nguyên)
nên \(x\in\left\{-3;0;3;6;9\right\}\)
=>G={-3;0;3;6;9}
a) \(2x^3-3x^2-5x=0\)
\(x\left(x+1\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=-1\left(TM\right)\\x=\dfrac{5}{2}\left(L\right)\end{matrix}\right.\)
\(A=\left\{-1\right\}\)
b) \(x< \left|3\right|\)\(\Leftrightarrow-3< x< 3\)
\(B=\left\{-2;-1;1;2\right\}\)
c) \(C=\left\{-3;3;6;9\right\}\)
a) \(A=\left\{x\in Z|2x^3-3x^2-5x=0\right\}\)
\(2x^3-3x^2-5x=0\)
\(\Leftrightarrow x\left(2x^2-3x-5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow A=\left\{0;-1\right\}\)
b) \(B=\left\{-2;-1;0;1;2\right\}\)
c) \(C=\left\{-3;3;6;9\right\}\)
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: \(D=\varnothing\)
Bạn ghi lại đề đi bạn. Với lại cho mình hỏi là đề bài yêu cầu gì vậy?
\(G=\left\{1;4;7;10\right\}\\ H=\left\{3\right\}\)