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\(A=\left\{x\in R|\left(x-2x^2\right)\left(x^2-3x+2\right)=0\right\}\)
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 \(B=\left(\dfrac{-1-5\sqrt[]{5}}{2};\dfrac{-1-\sqrt[]{13}}{2}\right)\cup\left(\dfrac{-1+\sqrt[]{13}}{2};\dfrac{-1+5\sqrt[]{5}}{2}\right)\)
\(\Rightarrow A\cap B=\left\{2\right\}\)
(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
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: \(\left(A\cap B\right)\cap C=\varnothing\)
\(\left(A\cup C\right)\cap\left(B\A\right)\)
\(=(-\dfrac{7}{4};-\dfrac{1}{2}]\cup\left(\dfrac{2}{3};+\infty\right)\cap\left[\left(-\dfrac{9}{2};-4\right)\cup\left(4;\dfrac{9}{2}\right)\right]\)
\(=\left(4;\dfrac{9}{2}\right)\)
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, \(A\cup B=(-5;2]\cup(3;6]\)
\(A\cap B=\left\{0\right\}\cup[4;5)\)
\(A\backslash B=(0;2]\cup\left[-5;6\right]\)
\(B\backslash A=[-5;0)\cup\left(3;4\right)\)
Bài 1:
\(|x-1|>3\Leftrightarrow \left[\begin{matrix} x-1>3\\ x-1< -3\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x>4\\ x< -2\end{matrix}\right.\)
\(\Rightarrow A=\left\{x\in\mathbb{R}|x\in (4;+\infty) \text{hoặc }x\in (-\infty;-2)\right\}\)
\(|x+2|< 5\Leftrightarrow -5< x+2< 5\Leftrightarrow -7< x< 3\Leftrightarrow x\in (-7;3)\)
\(\Rightarrow B=\left\{x\in\mathbb{R}|x\in (-7;3)\right\}\)
Do đó: \(A\cap B=\left\{\in\mathbb{R}|x\in (-7;-2)\right\}\)
Bài 2:
\(2< |x|\Leftrightarrow \left[\begin{matrix} x>2\\ x< -2\end{matrix}\right.(1)\)
\(|x|< 3\Leftrightarrow -3< x< 3(2)\)
Từ (1);(2) suy ra để $2< |x|< 3$ thì: \(\left[\begin{matrix} 2< x< 3\\ -3< x< -2\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x\in (2;3)\\ x\in (-3;-2)\end{matrix}\right.\)
Biểu diễn A qua hợp các khoảng:
\(A=(-3;-2)\cup (2;3)\)
A=(-2;2)
B=[-3;2)
A giao B=(-2;2)
A\B=\(\varnothing\)
B\A=[-3;-2]
\(C_R\left(A\cap B\right)=R\backslash\left(-2;2\right)=(-\infty;-2]\cup[2;+\infty)\)
\(A=\left[-3;3\right]\) ; \(B=(-\infty;-1]\cup[1;+\infty)\)
\(\Rightarrow A\cap B=\left[-3;-1\right]\cup\left[-1;3\right]\)