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(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
Nguyễn Huy TúAkai HarumaLightning FarronNguyễn Thanh HằngRibi Nkok NgokMysterious PersonVõ Đông Anh TuấnPhương AnTrần Việt Linh
\(A=\left\{x\in R|\left(x-2x^2\right)\left(x^2-3x+2\right)=0\right\}\)
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 \(B=\left(\dfrac{-1-5\sqrt[]{5}}{2};\dfrac{-1-\sqrt[]{13}}{2}\right)\cup\left(\dfrac{-1+\sqrt[]{13}}{2};\dfrac{-1+5\sqrt[]{5}}{2}\right)\)
\(\Rightarrow A\cap B=\left\{2\right\}\)
\(E=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A=\left\{1;-4\right\}\)
\(B=\left\{2;-1\right\}\)
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) \(A\cap B=\varnothing\)
\(\Rightarrow E\backslash\left(A\cap B\right)=\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)
\(A\cup B=\left\{-4;-1;1;2\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)=\left\{-5;-3;-2;0;3;4;5\right\}\)
\(\Rightarrow E\backslash\left(A\cup B\right)\subset E\backslash\left(A\cap B\right)\)
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, \(A\cup B=(-5;2]\cup(3;6]\)
\(A\cap B=\left\{0\right\}\cup[4;5)\)
\(A\backslash B=(0;2]\cup\left[-5;6\right]\)
\(B\backslash A=[-5;0)\cup\left(3;4\right)\)
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: \(\left(A\cap B\right)\cap C=\varnothing\)
\(\left(A\cup C\right)\cap\left(B\A\right)\)
\(=(-\dfrac{7}{4};-\dfrac{1}{2}]\cup\left(\dfrac{2}{3};+\infty\right)\cap\left[\left(-\dfrac{9}{2};-4\right)\cup\left(4;\dfrac{9}{2}\right)\right]\)
\(=\left(4;\dfrac{9}{2}\right)\)
\(a,\)\(A=\left\{x\in R|x< 3\right\}\Rightarrow A=\left(\text{ -∞;3}\right)\)
\(B=\left\{-1;0;1;2;3;4;5\right\}\)
\(\Rightarrow A\cap B=\left\{-1;0;1;2\right\}\)
\(b,x=-1\Rightarrow y=1-2\left(-1\right)+m=m+3\)
\(x=1\Rightarrow y=1-2+m=m-1\)
\(\Rightarrow C=(m-1;m+3]\subset A\)
\(\Rightarrow C\subset A\Leftrightarrow m+3< 3\Leftrightarrow m< 0\)
\(\frac{\left|x\right|-1}{\left|x\right|}>0\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\\left|x\right|-1>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
\(\Rightarrow C=\left(-\infty;-1\right)\cup\left(1;+\infty\right)\)
\(\Rightarrow B\cap C=(1;2019]\)
\(\Rightarrow A\cap B\cap C=\varnothing\Leftrightarrow\left[{}\begin{matrix}a>2019\\a+2\le1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a>2019\\a\le-1\end{matrix}\right.\)