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\(A=\left[-10;15\right]\) ; \(B=[12;+\infty)\); \(C=(-\infty;-8]\cup[5;+\infty)\)
\(A\cap B=\left[12;15\right]\)
\(A\backslash C=\left(-8;5\right)\)
\(B\backslash A=\left(15;+\infty\right)\)
Tập C chắc bạn viết nhầm, \(x< -8\) mới đúng, chứ chẳng ai cho vô lý thế kia
\(A=\left[-1;5\right]\) ; \(B=[2;+\infty)\); \(C=\left(-\infty;-8\right)\cup[2;+\infty)\)
\(A\cap B=\left[2;5\right]\) ; \(A\cup C=\left(-\infty;-8\right)\cup[-1;+\infty)\)
\(A\backslash B=[-1;2)\) ; \(B\backslash C=\varnothing\)
ĐKXĐ:
a/ \(\left\{{}\begin{matrix}3x+4\ge0\\x-3\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{4}{3}\\x\ne3\end{matrix}\right.\)
b/ \(x^2-5x+6\ne0\Rightarrow\left(x-2\right)\left(x-3\right)\ne0\Rightarrow\left\{{}\begin{matrix}x\ne2\\x\ne3\end{matrix}\right.\)
c/ \(\left\{{}\begin{matrix}4-x^2\ge0\\\left(x-2\right)\left(x-3\right)\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-2\le x\le2\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\Rightarrow-2\le x< 2\)
d/ \(\left\{{}\begin{matrix}4-x\ge0\\2x-10\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le4\\x\ge5\end{matrix}\right.\) \(\Rightarrow x=\varnothing\)
Bài 3:
a: \(\left(-\infty;\dfrac{1}{3}\right)\cap\left(\dfrac{1}{4};+\infty\right)=\left(\dfrac{1}{4};\dfrac{1}{3}\right)\)
b: \(\left(-\dfrac{11}{2};7\right)\cup\left(-2;\dfrac{27}{2}\right)=\left(-\dfrac{11}{2};\dfrac{27}{2}\right)\)
c: \(\left(0;12\right)\text{\[}5;+\infty)=\left(0;5\right)\)
d: \(R\[ -1;1)=\left(-\infty;-1\right)\cup[1;+\infty)\)
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: 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: \(\left(A\cap B\right)\cap C=\varnothing\)
\(\left(A\cup C\right)\cap\left(B\A\right)\)
\(=(-\dfrac{7}{4};-\dfrac{1}{2}]\cup\left(\dfrac{2}{3};+\infty\right)\cap\left[\left(-\dfrac{9}{2};-4\right)\cup\left(4;\dfrac{9}{2}\right)\right]\)
\(=\left(4;\dfrac{9}{2}\right)\)
\(A=\left[3;8\right]\) ; \(B=[10;+\infty)\) ; \(C=(-\infty;3]\cup[7;+\infty)\)
\(A\cap B=\varnothing\) ; \(A\cup C=\left(-\infty;+\infty\right)\)
\(A\backslash B=A=\left[3;8\right]\) ; \(B\backslash C=\varnothing\)