\(G=\left\{X\inℤ|X=\frac{3k-2}{k+1},k\inℤ\right\}\)

\(G=\left\{2;4;-2;8\right\}\)

\(c,20=2^2\cdot5\\ 45=3^2\cdot5\\ ƯCLN\left(20,45\right)=5\\ \RightarrowƯC\left(20,45\right)=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\\ C=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(d,\left(6x^2-7x+1\right)\left(x^3-x\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6x-1\right)x\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{6}\\x=1\\x=-1\end{matrix}\right.\)

\(\Leftrightarrow D=\left\{-1;0;\dfrac{1}{6};1\right\}\)

Sửa: \(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{6}\\x=1\\x=-1\end{matrix}\right.\Leftrightarrow D=\left\{-1;0;\dfrac{1}{6};1\right\}\)

Đề thiếu rồi bạn

\(a,C=\left\{0;5;10;15;20;25;30\right\}\\ b,x^2+3x-4=0\\ \Leftrightarrow x^2-x+4x-4=0\\ \Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+4\right)=0\\ \Leftrightarrow x-1=0.hoặc.x+4=0\\ \Leftrightarrow x=1.hoặc.x=-4\\ Vậy:D=\left\{-4;1\right\}\)

\(\left(2x+1\right)\left(x^2+x-1\right)\left(2x^2-3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x^2+x-1=0\\2x^2-3x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\) (pt \(x^2+x-1=0\) ko có nghiệm hữu tỉ nên ko cần quan tâm)

\(A=\left\{-\dfrac{1}{2};\dfrac{1}{2};1\right\}\)

con cãm ơn ạ

a: A={0;1;4;...}

b: B={1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}

c: C=B(120)={0;120;...}

\(x^4-6x^2+8=0\\ \Leftrightarrow x^4-2x^2-4x^2+8=0\\ \Leftrightarrow x^2\left(x-2\right)-4\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)^2\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

\(F=\left\{-2;2\right\}\)

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 \(F=\left\{2;-2;\sqrt{2};-\sqrt{2}\right\}\)

Bạn ghi lại đề đi bạn. Với lại cho mình hỏi là đề bài yêu cầu gì vậy?

Sorry bn