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

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

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

\(G=\left\{1;4;7;10\right\}\\ H=\left\{3\right\}\)

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;...}

\(B=\left\{x\in N/0\cdot x+2015=2015\right\}\)

=>\(x\in\left\{0,1,2,3,...\right\}\)

\(B=\left\{0;1;2;3;...\right\}\)

a: A={x\(\in R\)|x^2+x-6=0 hoặc 3x^2-10x+8=0}

=>x^2+x-6=0 hoặc 3x^2-10x+8=0

=>(x+3)(x-2)=0 hoặc (x-2)(3x-4)=0

=>\(x\in\left\{-3;2;\dfrac{4}{3}\right\}\)

=>A={-3;2;4/3}

B={x\(\in\)R|x^2-2x-2=0 hoặc 2x^2-7x+6=0}

=>x^2-2x-2=0 hoặc 2x^2-7x+6=0

=>\(x\in\left\{1+\sqrt{3};1-\sqrt{3};2;\dfrac{3}{2}\right\}\)

=>\(B=\left\{1+\sqrt{3};1-\sqrt{3};2;\dfrac{3}{2}\right\}\)

A={-3;2;4/3}

b: \(B\subset X;X\subset A\)

=>\(B\subset A\)(vô lý)

Vậy: KHông có tập hợp X thỏa mãn đề bài

D = {1; 2; 3; 4;...}

\(D=\left\{1;2;3;4;5;...\right\}\)

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