2+2*3+2+2+2+2+2+22+2+2+2+2+2+\(\infty\)
Những câu hỏi liên quan
1. Cho A[–4;7] và B(–infty;–2)∪ (3;+infty). Khi đó A∩B là:A) [–4;–2)∪ (3;7]B) [–4;–2)∪ (3;7).C) (–infty;2]∪ (3;+infty)D)(–infty;–2)∪ [3;+infty).2. Cho A(–infty;–2]; B[3;+infty) và C(0;4). Khi đó tập (A∪B)∩ C là:A) [3;4].B) (–infty;–2]∪ (3;+infty).C) [3;4).D)(–infty;–2)∪ [3;+infty).3. Cho A (−∞; 5), B (−∞; a) với a là số thực. Tìm a để A con BA. a 5.B. a ≤ 5.C. a ≥ 5.D. BA B
Đọc tiếp
1. Cho A=[–4;7] và B=(–\(\infty\);–2)∪ (3;+\(\infty\)). Khi đó A∩B là:
A) [–4;–2)∪ (3;7]
B) [–4;–2)∪ (3;7).
C) (–\(\infty\);2]∪ (3;+\(\infty\))
D)(–\(\infty\);–2)∪ [3;+\(\infty\)).
2. Cho A=(–\(\infty\);–2]; B=[3;+\(\infty\)) và C=(0;4). Khi đó tập (A∪B)∩ C là:
A) [3;4].
B) (–\(\infty\);–2]∪ (3;+\(\infty\)).
C) [3;4).
D)(–\(\infty\);–2)∪ [3;+\(\infty\)).
3. Cho A = (−∞; 5), B = (−∞; a) với a là số thực. Tìm a để A con B
A. a = 5.
B. a ≤ 5.
C. a ≥ 5.
D. B\A = B
tínha.limlimits_{n-+infty}dfrac{n^5+n^2-n+2}{left(2n^3-1right)left(n^2+n+1right)}b.limlimits_{n-+infty}dfrac{sqrt{n^2-n+2}}{n+2}c.limlimits_{n-+infty}dfrac{n-sqrt[3]{n^2-n^3}}{n^2+n+1}d.limlimits_{n-+infty}left(n-sqrt{n^2+n+1}right)
Đọc tiếp
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: \(\lim\limits_{n\rightarrow+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2-n^2-n-1}{n+\sqrt{n^2+n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-n-1}{n+\sqrt{n^2+n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-1-\dfrac{1}{n}}{1+\sqrt{1+\dfrac{1}{n}+\dfrac{1}{n^2}}}=-\dfrac{1}{1+1}=-\dfrac{1}{2}\)
Đúng 2
Bình luận (0)
tính
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}\)
3) \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}\)
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(3n+5-\dfrac{3}{n}\right)}{-n\left(1-\dfrac{5}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n+5-\dfrac{3}{n}}{-\left(1-\dfrac{5}{n}\right)}\)
\(=\left[{}\begin{matrix}-\infty\left(n\rightarrow+\infty\right)\\+\infty\left(n\rightarrow-\infty\right)\end{matrix}\right.\)
Bài 2,3 tương tự, bạn tự làm nhé!
Đúng 0
Bình luận (0)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2-2}+\sqrt[3]{x^3+1}}{\sqrt{x^2+1}-x}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x+3}{\sqrt{2x^2-3}}\)
\(\lim\limits_{x\rightarrow\pm\infty}\dfrac{2x^2-1}{3-x^2}\)
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\sqrt{\dfrac{4x^2}{x^2}-\dfrac{2}{x^2}}-x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{1}{x^3}}}{-x\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4}-1}{-1-1}=\dfrac{3}{2}\)
b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2x}{x}+\dfrac{3}{x}}{-\sqrt{\dfrac{2x^2}{x^2}-\dfrac{3}{x^2}}}=\dfrac{2}{-\sqrt{2}}=-\sqrt{2}\)
c/ \(\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}{\dfrac{3}{x^2}-\dfrac{x^2}{x^2}}=\dfrac{2}{-1}=-2\)
Đúng 1
Bình luận (0)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2-n+2}{n^3+2n^2-3}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{n+2}{3n^3-2n+n^2}\)
1: \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2-n+2}{n^3+2n^2-3}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1-\dfrac{1}{n}+\dfrac{2}{n^2}\right)}{n^3\left(1+\dfrac{2}{n}-\dfrac{3}{n^3}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1-\dfrac{1}{n}+\dfrac{2}{n^2}}{n\left(1+\dfrac{2}{n}-\dfrac{3}{n^3}\right)}=\lim\limits_{n\rightarrow\infty}\dfrac{1}{n}=0\)
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{n+2}{3n^3+n^2-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(1+\dfrac{2}{n}\right)}{n^3\left(3+\dfrac{1}{n}-\dfrac{2}{n^2}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^2}=0\)
Đúng 1
Bình luận (0)
tìm giới hạn của dãy số
1.\(\lim\limits_{n->\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)
2.\(\lim\limits_{n->\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)
3.tìm a,b để \(\lim\limits_{n->\infty}\left(\sqrt{an^2+bn+2}-2n\right)=2\)
1: \(\lim\limits_{n\rightarrow\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^3+n^2+n+1-n^3}{\sqrt[3]{\left(n^3+n^2+n+1\right)^2}+n\cdot\sqrt[3]{n^3+n^2+n+1}+n^2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n+1}{n^2\cdot\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+n^2\cdot\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n^2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{1}{n}+\dfrac{1}{n^2}}{\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+1}\)
\(=\dfrac{1}{1+1+1}=\dfrac{1}{3}\)
2: \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n-n^2+n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2-\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}}\)
\(=\dfrac{2}{1+1}=\dfrac{2}{2}=1\)
Đúng 1
Bình luận (0)
tìm tập xác định của hàm số y=\(\sqrt{-2x^2+5x-2}\)
A. D=\((-\infty;\dfrac{1}{2}]\) B. \([2;+\infty)\) C. \((-\infty;\dfrac{1}{2}]\)u\([2;+\infty)\) D. \([\dfrac{1}{2};2]\)
giải chi tiết
H/s xác định `<=>-2x^2+5x-2 >= 0`
`<=>1/2 <= x <= 2`
`->\bb D`
Đúng 6
Bình luận (0)
\(\sin^2\infty-2\sin^2\infty=\frac{1}{4}\)
\(7\sin^2\infty-5\cos^2\infty=6.5\)
\(\sin\left(90-\infty\right)-3\cos\infty=1.5\)
tính số đo góc nhọn \(\infty\)
1. Cho A (1; +∞); B [−2; 6] . Tập hợp A ∩ B làA. [−2; +∞)B. (1; +∞)C. [−2; 6]D. (1; 6]2. Cho A[–4;7] và B(-infty;–2)∪ (3;+infty). Khi đó A∩B là:A.[– 4; – 2) ∪ (3; 7] B.[– 4; – 2) ∪ (3; 7)C.(– ∞; 2] ∪ (3; +∞)D.(−∞; −2) ∪ [3; +∞)3. Cho ba tập hợp A (-∞; 3), B [−1; 8], C (1 ; +∞). Tập (A ∩ B) (A ∩ C) là tậpA. [−1; 1] B. (1 ; 3)C. (−1; 3)D. (−1; 1)
Đọc tiếp
1. Cho A = (1; +∞); B = [−2; 6] . Tập hợp A ∩ B là
A. [−2; +∞)
B. (1; +∞)
C. [−2; 6]
D. (1; 6]
2. Cho A=[–4;7] và B=(-\(\infty\);–2)∪ (3;+\(\infty\)). Khi đó A∩B là:
A.[– 4; – 2) ∪ (3; 7]
B.[– 4; – 2) ∪ (3; 7)
C.(– ∞; 2] ∪ (3; +∞)
D.(−∞; −2) ∪ [3; +∞)
3. Cho ba tập hợp A = (-∞; 3), B = [−1; 8], C = (1 ; +∞). Tập (A ∩ B)\ (A ∩ C) là tập
A. [−1; 1]
B. (1 ; 3)
C. (−1; 3)
D. (−1; 1)
