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a/ \(\left\{a\right\};\left\{b\right\};\left\{a;b\right\};\varnothing\)
b/ \(\left\{1\right\};\left\{2\right\};\left\{3\right\};\left\{1;2\right\};\left\{1;3\right\};\left\{2;3\right\};\left\{1;2;3\right\};\varnothing\)
c/ \(\left\{0\right\};\left\{1\right\};\left\{2\right\};\left\{3\right\};\left\{0;1\right\};\left\{0;2\right\};\left\{0;3\right\};\left\{1;2\right\};\left\{1;3\right\};\left\{2;3\right\};\left\{0;1;2\right\};\left\{1;2;3\right\};\left\{0;2;3\right\};\left\{0;1;3\right\};\left\{0;1;2;3\right\};\varnothing\)
d/ \(\left\{1\right\};\left\{-2\right\};\left\{1;-2\right\};\varnothing\)
Câu 2:
\(X\subset\left\{-3;-2;0;1;2;3\right\}\)
\(X\subset\left\{-1;0;1;2;3;4\right\}\)
DO đó: \(X=\left\{0;1;2;3\right\}\)
Các tập con là {0}; {1}; {2}; {3}; rỗng; {0;1}; {0;2}; {0;3}; {1;2}; {1;3}; {2;3}; {0;1;2}; {1;2;3}; {0;1;3}; {0;1;2;3}
\(\left(x-1\right)\left(x+2\right)\left(x^3+4x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)x\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-2\notin N\left(l\right)\end{matrix}\right.\)
\(\Rightarrow A=\left\{0;1\right\}\)
Các tập con của A là: \(\varnothing;\left\{0\right\};\left\{1\right\};\left\{0;1\right\}\)
1/ B={x ∈ R| (9-x2)(x2-3x+2)=0}
Ta có:
(9-x2)(x2-3x+2)=0
⇔\(\left[{}\begin{matrix}9-x^2=0\\x^2-3x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(3+x\right)\left(3-x\right)=0\\\left(x^2-x\right)-\left(2x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x\left(x-1\right)-2\left(x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\\left(x-1\right)\left(x-2\right)=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=1\\x=2\end{matrix}\right.\)
⇒B={-3;1;2;3}
2/ Có 15 tập hợp con có 2 phần tử
a) \(A=\left\{0;1;2;3;...;13\right\}\)
b) Ta có: \(x^2+3x-9=0\)
\(\Leftrightarrow\left(x-\frac{-3+3\sqrt{5}}{2}\right)\left(x+\frac{3+3\sqrt{5}}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-3+3\sqrt{5}}{2}\\x=\frac{-3-3\sqrt{5}}{2}\end{cases}}\)
c) \(C=\left\{-7;-6;-5;...;5;6;7\right\}\)