a) (-\(\infty\);\(\dfrac{1}{3}\))\(\cap\)(\(\dfrac{1}{4}\);+\(\infty\))
b)\(\left(-\dfrac{11}{2};7\right)\cap\left(-2;\dfrac{27}{2}\right)\)
c)(0;12) \ [5;+\(\infty\))
d) R\[-1;1)
mọi người giúp em với ạ
Bài 3: Tìm giao các tập hợp sau:
\(a,\left(-\infty;\dfrac{1}{3}\right)\cap\left(\dfrac{1}{4};+\infty\right)\\ b,\left(-\dfrac{11}{2};7\right)\cap\left(-2;\dfrac{27}{2}\right)\\ c,\left(0;12\right)\cap[5;+\infty)\\ d,R\cap[-1;1)\)
\(\left(-\infty;\dfrac{1}{3}\right)\cap\left(\dfrac{1}{4};+\infty\right)=\left(\dfrac{1}{4};\dfrac{1}{3}\right)\)
\(\left(-\dfrac{11}{2};7\right)\cap\left(-2;\dfrac{27}{2}\right)=\left(-2;7\right)\)
\(\left(0;12\right)\cap[5;+\infty)=[5;12)\)
\(R\cap\left[-1;1\right]=\left[-1;1\right]\)
Bài 1: Xác định A\(\cap\)B, A\(\cup\)B, A\B, B/A và biểu diễn kết quả trên trục số
a, A = {\(x\in\) R |x \(\ge\) 1} B = {\(x\in\) R |x \(\le\) 3}
b, A = {\(x\in\) R |x \(\le\) 1} B = {\(x\in\) R |x \(\ge\) 3}
c, A = [1;3] B = (2;+\(\infty\))
d, A = (-1;5) B = [0;6)
Bài 2: Cho A = {\(x\in\) R |x - 2 \(\ge\) 0}, B = {\(x\in\) R |x - 5 > 0}
Tính A\(\cap\)B, A\(\cup\)B, A\B, B\A
Bài 3: Xác định các tập sau
a, \(\left(-\infty;\dfrac{1}{3}\right)\cap\left(\dfrac{1}{4};+\infty\right)\)
b, \(\left(-\dfrac{11}{2};7\right)\cup\left(-2;\dfrac{27}{2}\right)\)
c, (0;12) \ [5;+\(\infty\))
d, R \ [-1;1)
Gíup với ạ!!!
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)\)
Cho \(A=\left[m-1;\dfrac{m+3}{2}\right]\); \(B=\left(-\infty;-3\right)\cup[3;+\infty)\)
Tìm m để \(A\cap B\ne\varnothing\)
Dễ thấy nếu \(A\cap B=\varnothing\Rightarrow A\in[-3;3)\Rightarrow\left\{{}\begin{matrix}m-1\ge-3\\\dfrac{m+3}{2}< 3\end{matrix}\right.\)
\(\Leftrightarrow-2\le m< 3\)
Do đó để \(A\cap B\ne\varnothing\Rightarrow m\notin[-2;3)\Rightarrow\left[{}\begin{matrix}m< -2\\m\ge3\end{matrix}\right.\)
tính
a.\(\lim\limits_{n->+\infty}\dfrac{n^5+n^2-n+2}{\left(2n^3-1\right)\left(n^2+n+1\right)}\)
b.\(\lim\limits_{n->+\infty}\dfrac{\sqrt{n^2-n+2}}{n+2}\)
c.\(\lim\limits_{n->+\infty}\dfrac{n-\sqrt[3]{n^2-n^3}}{n^2+n+1}\)
d.\(\lim\limits_{n->+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
a: \(\lim\limits_{n\rightarrow+\infty}\dfrac{n^5+n^2-n+2}{\left(2n^3-1\right)\left(n^2+n+1\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n^3}-\dfrac{1}{n^4}+\dfrac{2}{n^5}}{\left(\dfrac{2n^3}{n^3}-\dfrac{1}{n^3}\right)\left(\dfrac{n^2+n+1}{n^2}\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n^3}-\dfrac{1}{n^4}+\dfrac{2}{n^5}}{\left(2-\dfrac{1}{n^3}\right)\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}\)
\(=\dfrac{1}{2\cdot1}=\dfrac{1}{2}\)
b: \(\lim\limits_{n\rightarrow+\infty}\dfrac{\sqrt{n^2-n+2}}{n+2}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n\sqrt{1-\dfrac{1}{n}+\dfrac{2}{n^2}}}{n\left(1+\dfrac{2}{n}\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\sqrt{1-\dfrac{1}{n}+\dfrac{2}{n^2}}}{1+\dfrac{2}{n}}=\dfrac{\sqrt{1-0+0}}{1+0}=\dfrac{1}{1}=1\)
c: \(\lim\limits_{n\rightarrow+\infty}\dfrac{n-\sqrt[3]{n^2-n^3}}{n^2+n+1}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\dfrac{n}{n^2}-\dfrac{\sqrt[3]{n^2-n^3}}{n^2}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\dfrac{1}{n}-\sqrt[3]{\dfrac{1}{n^4}-\dfrac{1}{n^3}}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}=\dfrac{0}{1}=0\)
d: \(\lim\limits_{n\rightarrow+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2-n^2-n-1}{n+\sqrt{n^2+n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-n-1}{n+\sqrt{n^2+n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-1-\dfrac{1}{n}}{1+\sqrt{1+\dfrac{1}{n}+\dfrac{1}{n^2}}}=-\dfrac{1}{1+1}=-\dfrac{1}{2}\)
Xác định các tập hợp sau và biểu diễn chúng trên trục số :
a) ( \(-\infty;3\)] \(\cap\left(-2;+\infty\right)\)
b) \(\left(-15;7\right)\cup\left(-2;14\right)\)
c) \(\left(0;12\right)\)\ [ \(5;+\infty\))
d) \(R\)\ \(\left(-1;1\right)\)
a) (\(-2;3\)]
b) \(\left(-15;14\right)\)
c) \(\left(0;5\right)\)
d) (\(-\infty;4\)] \(\cup\) [\(1;+\infty\))
a) (−∞;3]∩(−2;+∞)=(−2;3](−∞;3]∩(−2;+∞)=(−2;3]
b) (0;12)∩[5;+∞)=(0;5)(0;12)∩[5;+∞)=(0;5)
c) (−15,7)∪(−2;14)=(−2;1)∪(3;7)(−15,7)∪(−2;14)=(−2;1)∪(3;7)
d) R∖(−1;1)=(−∞;−1]∪[1;+∞)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{3x^3+1}-\sqrt{2x^2+x+1}}{\sqrt[4]{4x^4+2}}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(2x+1\right)^3\left(x+2\right)^4}{\left(3-2x\right)^7}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{4x^2-3x+4}-2x}{\sqrt{x^2+x+1}-x}\)
Da nan roi mang meo lam mat het bai -.-
1/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{\dfrac{3x^3}{x^3}+\dfrac{1}{x^3}}+\sqrt{\dfrac{2x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}{-\sqrt[4]{\dfrac{4x^4}{x^4}+\dfrac{2}{x^4}}}=\dfrac{-\sqrt[3]{3}-\sqrt{2}}{\sqrt[4]{4}}\)
2/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{8x^7}{\left(-2x^7\right)}=-\dfrac{8}{2^7}\)
3/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(4x^2-3x+4-4x^2\right)\left(\sqrt{x^2+x+1}+x\right)}{\left(x^2+x+1-x^2\right)\left(\sqrt{4x^2-3x+4}+2x\right)}=\dfrac{-3.2}{2}=-3\)
Tập nghiệm của bất phương trình \(\left|x+1\right|\)<x là:
A. \(S=\left(\dfrac{1}{2};+\infty\right)\) B. \(S=\left(0;\dfrac{1}{2}\right)\) C. \(S=\varnothing\) D. \(S=\left(-\infty;-\dfrac{1}{2}\right)\)
1. Tìm GTNN của \(y=x+\dfrac{1}{x}-5\) trên \(\left(0,+\infty\right)\)
2. Tìm GTNN của \(y=4x^2+\dfrac{1}{x}-4\) trên \(\left(0,+\infty\right)\)
3. Tìm GTLN của \(y=\dfrac{x^2+4}{x}\) trên \(\left(-\infty,0\right)\)
\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)
\(y_{min}=-3\) khi \(x=1\)
\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)
\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)
\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)
\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)
Tìm lim (\(\dfrac{2021}{n^2}-\left(\dfrac{3}{7}\right)^n+2022\))
A. 2022 B.0 c.\(\infty\) d.-\(\infty\)