Tìm \(n\in N\), sao cho :
\(a,\left(2n^2-3n+1\right)⋮\left(n-1\right)\)
\(b,\left(2n^2-3n+1\right)⋮\left(2n-1\right)\)
Tìm các giới hạn sau:
a) \(lim\left(4^n-3^n\right)\)
b) \(lim\left[\left(2^n+1\right)^2-4^n\right]\)
c) \(lim\left(\sqrt{2n^5-3n^2+11}-n^3\right)\)
d) \(lim\left(\sqrt{2n^2+1}-\sqrt{3n^2-1}\right)\)
e) \(lim\sqrt{n^2+3n\sqrt{n}+1}-n\)
\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)
\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)
\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)
\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)
Tìm các giới hạn sau:
\(a,\dfrac{\left(2n+1\right)\left(3n-2\right)^2}{n^3+n-1}\)
\(b,\dfrac{2n-1}{3n^2+4n-1}\)
\(\lim\dfrac{\left(2n+1\right)\left(3n-2\right)^2}{n^3+n-1}=\lim\dfrac{n\left(2+\dfrac{1}{n}\right).n^2.\left(3-\dfrac{2}{n}\right)^2}{n^3\left(1+\dfrac{1}{n^2}-\dfrac{1}{n^3}\right)}\)
\(=\lim\dfrac{\left(2+\dfrac{1}{n}\right)\left(3-\dfrac{2}{n}\right)^2}{1+\dfrac{1}{n^2}-\dfrac{1}{n^3}}=\dfrac{2.3^2}{1}=18\)
\(\lim\dfrac{2n-1}{3n^2+4n-1}=\lim\dfrac{n\left(2-\dfrac{1}{n}\right)}{n^2\left(3+\dfrac{4}{n}-\dfrac{1}{n^2}\right)}=\lim\dfrac{2-\dfrac{1}{n}}{n\left(3+\dfrac{4}{n}-\dfrac{1}{n^2}\right)}=\dfrac{2}{+\infty}=0\)
tính các giới hạn sau:
a) lim (3n2+n2-1)
b)lim \(\dfrac{n^3+3n+1}{2n-n^3}\)
c) lim \(\dfrac{-2n^3+3n+1}{n-n^2}\)
d) lim \(\left(n+\sqrt{n^2-2n}\right)\)
e) lim \(\left(2n-3.2^n+1\right)\)
f) lim \(\left(\sqrt{4n^2-n}-2n\right)\)
g) lim \(\left(\sqrt{n^2+3n-1}-\sqrt[3]{n^3-n}\right)\)
a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả
b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)
c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)
d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)
e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)
f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)
g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)
\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)
\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)
a) lim \(\left(-3n^3+n^2-1\right)\)
minh le oi ban dao mau so cua ban len cho tu uong roi thay vi tri cua mau thanh n3 +2n
Chứng minh rằng với mọi n thuộc Z thì :
a) \(\left(n^2+3n-1\right).\left(n+2\right)-n^3+2⋮5\)
b) \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)⋮2\)
c) \(\left(2n-1\right).3-\left(2n-1\right)⋮8\)
d) \(n^2\left(n+1\right)+2n\left(n+1\right)⋮6\)
a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10⋮2\)
d: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
Bài 1: CMR
a) A = \(\frac{\left(n+1\right).\left(n+2\right)....\left(2n-1\right).\left(2n\right)}{2^n}\) là số nguyên.
b) B = \(\frac{3.\left(n+1\right).\left(n +2\right)...\left(3n-1\right).3n}{3^n}\)là số nguyên.
Tính giới hạn :
L = lim \(\dfrac{\left(n^2+2n\right)\left(2n^3+1\right)\left(4n+5\right)}{\left(n^4-3n-1\right)\left(3n^2-7\right)}\)
Dang này thì cứ chọn số hạng có mũ cao nhất trên tử và mẫu là được. Nó là ngắt vô cùng lớn hay bé gì đấy
\(=lim\dfrac{8n^6}{3n^6}=\dfrac{8}{3}\)
a; lim\(\frac{\sqrt{6n^4+n+1}}{2n^2+1}\)
b; lim \(\frac{\left(n+1\right)\left(2n+1\right)^2\left(3n+1\right)^3}{n^2\left(n+2\right)^2\left(1-3n\right)^2}\)
\(1,\left(n+2\right)⋮\left(n+1\right)\)
2 ,\(8⋮\left(n-2\right)\)
3,\(\left(2n+1\right)⋮\left(6-n\right)\)
4;\(3n⋮\left(n-1\right)\)
5, \(\left(3n+5\right)⋮\left(2n+1\right)\)
6, \(\left(3n+1\right)⋮\left(2n-1\right)\)
Tìm các giới hạn sau:
a) \(lim\left(\sqrt{4n+1}-2\sqrt{n}\right)\)
b) \(lim\left(\sqrt{n^2+2n}-\sqrt{n^2-2n}-n\right)\)
c) \(lim\left(\sqrt{9^n-3^n}-4^n\right)\)
d) \(lim\left(3n^3+2n^2+n\right)\)
\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)
\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)=+\infty.3=+\infty\)