3. Cho A=\(\left\{x^{ }\in R|\left(x^2-7x+6\right)\left(x^2-4\right)=0\right\}\)
B=\(\left\{x\in Z|-3< x< \sqrt{7}\right\}\)
Tìm A\(\cap\)B, A\(\cup\)B
Cho \(E=\left\{x\in Z|\left|x\right|\le5\right\}\); \(A=\left\{x\in R|x^2+3x-4=0\right\}\);
\(B=\left\{x\in Z|(x-2)(x+1)(2x^2-x-3)=0\right\}\)
a) CM \(A\subset E\),\(B\subset E\)
b) Tìm \(E\backslash\left(A\cap B\right)\),\(E\backslash\left(A\cup B\right)\) rồi tìm quan hệ giữa hai tập hợp này.
\(E=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A=\left\{1;-4\right\}\)
\(B=\left\{2;-1\right\}\)
a) Với mọi x thuộc A đều thuộc E \(\Rightarrow A\subset E\)
Với mọi x thuộc B đều thuộc E \(\Rightarrow B\subset E\)
b) \(A\cap B=\varnothing\)
\(\Rightarrow E\backslash\left(A\cap B\right)=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A\cup B=\left\{-4;-1;1;2\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)=\left\{-5;-3;-2;0;3;4;5\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)\subset E\backslash\left(A\cap B\right)\)
Cho \(A=\left\{x\in R|x^2-7x+6=0\right\}\)
\(B=\left\{x\in Z|\left|x\right|< 4\right\}\)
Xác định là tập hợp : \(A\cup B;A\cap B;\) A \B: B \ A
\(A=\left\{1;6\right\}\) ; \(B=\left(-4;4\right)\)
\(A\cup B=\left(-4;4\right)\cup\left\{6\right\}\)
\(A\cap B=\left\{1\right\}\)
\(A\backslash B=\left\{6\right\}\)
\(B\backslash A=\left(-4;1\right)\cup\left(1;4\right)\)
Xác định các tập: \(A\cup B,A\cap B;A\backslash B;B\backslash A\)
a, \(A=\left\{x\in R|-3\le x\le5\right\};B==\left\{x\in R|\left|x\right|< 4\right\}\)
b, \(A=\left[1;5\right];B=\left(-3;2\right)\cup\left(3;7\right)\)
c, \(A=\left\{x\in R|\dfrac{1}{\left|x-1\right|}\ge2\right\};B=\left\{x\in R|\left|x-2\right|\le1\right\}\)
d, \(A=\left[0;2\right]\cup\left(4;6\right);B=(-5;0]\cup\left(3;5\right)\)
a, \(A\cup B=(-4;5]\)
\(A\cap B=[-3;4)\)
\(A\backslash B=\left[4;5\right]\)
\(B\backslash A=\left(-4;-3\right)\)
b, \(A\cup B=\left(-3;7\right)\)
\(A\cap B=[1;2)\cup(3;5]\)
\(A\backslash B=\left[2;3\right]\)
\(B\backslash A=\left(-3;1\right)\cup\left(5;7\right)\)
c, \(A\cup B=\left[\dfrac{1}{2};3\right]\)
\(A\cap B=\left[1;\dfrac{3}{2}\right]\)
\(A\backslash B=[\dfrac{1}{2};1)\)
\(B\backslash A=(\dfrac{3}{2};3]\)
d, \(A\cup B=(-5;2]\cup(3;6]\)
\(A\cap B=\left\{0\right\}\cup[4;5)\)
\(A\backslash B=(0;2]\cup\left[-5;6\right]\)
\(B\backslash A=[-5;0)\cup\left(3;4\right)\)
Tìm \(A\cap B;A\cup B;A/B;B/A\)
\(A=\left\{x\in Z|x^2< 4\right\};B=\left\{x\in Z|\left(5x-3x^2\right)\left(x^2-2x-3\right)=0\right\}
\)
\(A=\left\{x\in Z,x^2< 4\right\}\)
\(\Rightarrow A=\left\{-1;0;1\right\}\)
\(B=\left\{x\in Z,\left(5x-3x^2\right)\left(x^2-2x-3\right)=0\right\}\)\(\Rightarrow\left[{}\begin{matrix}5x-3x^2=0\\x^2-2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{3}\left(loai\right)\\x=0\\x=3\\x=-1\end{matrix}\right.\)
\(\Rightarrow B=\left\{0;-1;3\right\}\)
\(\Rightarrow A\cap B=\left\{0;-1\right\}\) \(A\cup B=\left\{0;-1;1;3\right\}\)
\(A\backslash B=\left\{1\right\}\) \(B\backslash A=\left\{3\right\}\)
cho các tập hợp sau:
A={x\(\in\)R|(2x-\(x^2\))(2\(x^3\)-3x-2)=0};B={n\(\in N\)*|3<\(n^2\)<30}
A. \(A\cap B=\left\{2;4\right\}\)
B. \(A\cap B=\left\{2\right\}\)
C. \(A\cap B=\left\{4;5\right\}\)
D. \(A\cap B=\left\{3\right\}\)
(2x-x^2)(2x^3-3x-2)=0
=>x(2-x)(2x^3-3x-2)=0
=>x=0 hoặc 2-x=0 hoặc 2x^3-3x-2=0
=>\(x\in\left\{0;2;1,48\right\}\)
=>\(A=\left\{0;2;1,48\right\}\)
3<n^2<30
mà \(n\in Z^+\)
nên \(n\in\left\{2;3;4;5\right\}\)
=>B={2;3;4;5}
=>A giao B={2}
=>Chọn B
Cho các tập hợp sau A= \(\left\{x\in R|\left(x-2x^2\right)\left(x^2-3x+2\right)=0\right\}\) và B=\(\left\{n\in N|3< n\left(n+1\right)< 31\right\}\)
Tìm A \(\cap\) B
\(A=\left\{x\in R|\left(x-2x^2\right)\left(x^2-3x+2\right)=0\right\}\)
Giải phương trình sau :
\(\left(x-2x^2\right)\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow x\left(1-2x\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1-2x=0\\x-1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\\x=2\end{matrix}\right.\)
\(\Rightarrow A=\left\{0;\dfrac{1}{2};1;2\right\}\)
\(B=\left\{n\in N|3< n\left(n+1\right)< 31\right\}\)
Giải bất phương trình sau :
\(3< n\left(n+1\right)< 31\)
\(\Leftrightarrow\left\{{}\begin{matrix}n\left(n+1\right)>3\\n\left(n+1\right)< 31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n^2+n-3>0\\n^2+n-31< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n< \dfrac{-1-\sqrt[]{13}}{2}\cup n>\dfrac{-1+\sqrt[]{13}}{2}\\\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1-\sqrt[]{13}}{2}\\\dfrac{-1+\sqrt[]{13}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)
Vậy \(B=\left(\dfrac{-1-5\sqrt[]{5}}{2};\dfrac{-1-\sqrt[]{13}}{2}\right)\cup\left(\dfrac{-1+\sqrt[]{13}}{2};\dfrac{-1+5\sqrt[]{5}}{2}\right)\)
\(\Rightarrow A\cap B=\left\{2\right\}\)
a, A = [ -2; 5)
B= ( - \(\infty\); 3 ]
C=(- \(\infty\) ; 4 )
Cho \(A=\left\{x\subseteq Z/x^2< \frac{15}{2}\right\}\)
\(B=\left\{0,1,3\right\},C=\left\{x\subseteq R/\left(2X-3\right)\left(X^2-4\right)=0\right\}\)
Tim \(A\cap\left(B\cup C\right)\)
\(A=\left\{-2;-1;0;1;2\right\}\)
\(C=\left\{-2;\frac{3}{2};2\right\}\)
\(\Rightarrow B\cup C=\left\{-2;0;1;\frac{3}{2};2;3\right\}\)
\(\Rightarrow A\cap\left(B\cup C\right)=\left\{-2;0;1;2\right\}\)
Cho tập hợp: A=\(\left\{x\in R:-\dfrac{7}{4}< x\le-\dfrac{1}{2}\right\}\), B=\(\left\{x\in R:4< \left|x\right|< \dfrac{9}{2}\right\}\),C=\(\left\{x\in R:-\dfrac{5}{2}x+3< 3x-\dfrac{2}{3}\right\}\)
a. Dùng kí hiệu đoạn, khoảng, nửa khoảng để viết lại các tập hợp trên.
b. Xác định \(\left(A\cap B\right)\)\(\cap C\), \(\left(CrA\right)\)trừ B, \(\left(A\cup C\right)\)\(\cap\)(B trừ A)
a: A=(-7/4; -1/2]
\(B=\left(-\dfrac{9}{2};-4\right)\cup\left(4;\dfrac{9}{2}\right)\)
\(C=\left(\dfrac{2}{3};+\infty\right)\)
b: \(\left(A\cap B\right)\cap C=\varnothing\)
\(\left(A\cup C\right)\cap\left(B\A\right)\)
\(=(-\dfrac{7}{4};-\dfrac{1}{2}]\cup\left(\dfrac{2}{3};+\infty\right)\cap\left[\left(-\dfrac{9}{2};-4\right)\cup\left(4;\dfrac{9}{2}\right)\right]\)
\(=\left(4;\dfrac{9}{2}\right)\)