Tìm \(n\in N\)để:
a/ \(\left(n+8\right)⋮n\)
b/ \(5⋮\left(n+3\right)\)
c/ \(\left(n+8\right)⋮\left(n+3\right)\)
d/ \(3n+13⋮n+2\)
e/ \(\left(5n+2\right)⋮\left(9-2n\right)\)
a) \(lim\frac{\left(-2\right)^n+3^n}{\left(-2\right)^{n+1}+3^{n+1}}\)
b) \(lim\frac{\left(2n-1\right)\left(n+1\right)\left(3n+4\right)}{\left(5-6n\right)^3}\)
c) \(lim\left(\sqrt{n^2+5n+1}-\sqrt{n^2-2}\right)\)
d) \(lim\frac{5\cdot3^n-6^{n+1}}{4\cdot2^n+6^n}\)
e) \(lim\left(-2n^3-3n^2+5n-2020\right)\)
a/ \(=lim\frac{\left(-\frac{2}{3}\right)^n+1}{-2.\left(-\frac{2}{3}\right)^n+3}=\frac{1}{3}\)
b/ \(=lim\frac{\left(2-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(3+\frac{4}{n}\right)}{\left(\frac{5}{n}-6\right)^3}=\frac{2.1.3}{\left(-6\right)^3}=-\frac{1}{36}\)
c/ \(=lim\frac{5n+3}{\sqrt{n^2+5n+1}+\sqrt{n^2-2}}=\frac{5+\frac{3}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{2}{n}}}=\frac{5}{1+1}=\frac{5}{2}\)
d/ \(=lim\frac{5.\left(\frac{1}{2}\right)^n-6}{4.\left(\frac{1}{3}\right)^n+1}=\frac{-6}{1}=-6\)
e/ \(=-n^3\left(2+\frac{3}{n}-\frac{5}{n^2}+\frac{2020}{n^3}\right)=-\infty.2=-\infty\)
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) \(lim\sqrt{n^2+3n\sqrt{n}+1}-n\)
\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)
\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)
\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)
\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)
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: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
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) \(lim\left(3n^3+2n^2+n\right)\)
\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)
\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)=+\infty.3=+\infty\)
Tìm các số nguyên n thỏa mãn :
a)\(\left(n+5\right)⋮\left(n-2\right)\)
b)\(\left(2n+1\right)⋮\left(n-5\right)\)
c) \(\left(n^2+3n-13\right)⋮n+3\)
d)\(\left(n^2+3\right)⋮\left(n-1\right)\)
tìm : \(n\in N\)để :
a)\(\left(2n+9\right)⋮\left(3n+1\right)\)
b)\(\left(5n+2\right)⋮\left(9-2n\right)\)
\(2n+9⋮3n+1\)
\(\Rightarrow3\left(2n+9\right)⋮3n+1\)
\(\Rightarrow2\left(3n+1\right)+25⋮3n+1\)
\(\Rightarrow25⋮3n+1\)
\(\Rightarrow3n+1\in\left\{5,25,1,-5,-25,-1\right\}\)
\(n\in\left\{8,0\right\}\)
\(5n+2⋮9-2n\)
\(\Rightarrow2\left(5n+2\right)⋮9-2n\)
\(\Rightarrow-5\left(9-2n\right)-41⋮9-2n\)
\(41⋮9-2n\)
\(\Rightarrow9-2n\in\left\{41,-41,1,-1\right\}\)
\(\Rightarrow n\in\left\{-16,25,4,-5\right\}\)
\(1,\left(n+2\right)⋮\left(n+1\right)\)
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, \(\left(3n+1\right)⋮\left(2n-1\right)\)
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
tìm các giới hạn
a)lim(\(\sqrt{n+1}-\sqrt{n}\))
b)lim\(\left(\sqrt{n+5n+1}-\sqrt{n^2-n}\right)\)
c)lim\(\left(\sqrt{3n^2+2n-1}-\sqrt{3n^2-4n+8}\right)\)
d)lim\(\frac{2^n+6^n-4^{n+1}}{3^n+6^{n+1}}\)
e)lim\(\frac{3^n-4^n+5^n}{3^n+4^n-5^n}\)
f)lim\(\frac{1+3+5+.....+\left(2n+1\right)}{3n^2+4}\)
g)lim[\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{n\left(n+1\right)}\)]
h)lim\(\frac{1^2+2^2+3^2+.....+n^2}{n\left(n+1\right)\left(n+2\right)}\)
a/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)
b/ \(=lim\frac{6n+1}{\sqrt{n^2+5n+1}+\sqrt{n^2-n}}=\frac{6+\frac{1}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}=\frac{6}{1+1}=3\)
c/ \(=lim\frac{6n-9}{\sqrt{3n^2+2n-1}+\sqrt{3n^2-4n+8}}=lim\frac{6-\frac{9}{n}}{\sqrt{3+\frac{2}{n}-\frac{1}{n^2}}+\sqrt{3-\frac{4}{n}+\frac{8}{n^2}}}=\frac{6}{\sqrt{3}+\sqrt{3}}=\sqrt{3}\)
d/ \(=lim\frac{\left(\frac{2}{6}\right)^n+1-4\left(\frac{4}{6}\right)^n}{\left(\frac{3}{6}\right)^n+6}=\frac{1}{6}\)
e/ \(=lim\frac{\left(\frac{3}{5}\right)^n-\left(\frac{4}{5}\right)^n+1}{\left(\frac{3}{5}\right)^n+\left(\frac{4}{5}\right)^n-1}=\frac{1}{-1}=-1\)
f/ Ta có công thức:
\(1+3+...+\left(2n+1\right)^2=\left(n+1\right)^2\)
\(\Rightarrow lim\frac{1+3+...+2n+1}{3n^2+4}=lim\frac{\left(n+1\right)^2}{3n^2+4}=lim\frac{\left(1+\frac{1}{n}\right)^2}{3+\frac{4}{n^2}}=\frac{1}{3}\)
g/ \(=lim\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\right)=lim\left(1-\frac{1}{n+1}\right)=1-0=1\)
h/ Ta có: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)