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 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: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
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.
Tìm tất cả các số nguyên n để:
a) \(\left(2n^2+n-7\right)\) ⋮\(\left(n-2\right)\)
b) \(\left(10n^2+n-10\right)\) ⋮\(\left(n-1\right)\)
c) \(\left(3n^3+10n^2-5\right)\) ⋮\(\left(3n+1\right)\)
d) \(\left(n^3-3n^2-3n-1\right)\)⋮\(\left(n^2+n+1\right)\)
Lan nghĩ ra một số biết rằng số đó bằng hiệu của số chẵn lớn nhất có 3 chữ số chẵn khác nhau với 60 rồi cộng thêm 21. Hỏi số lan nghĩ là số nào
Tìm các giới hạn sau:
\(a,\dfrac{4n^5-3n^2}{\left(3n^2-2\right)\left(1-4n^3\right)}\)
\(b,\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)
\(b,lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)
\(=lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(\dfrac{1}{n}-\dfrac{10}{n^2}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(\dfrac{3}{n^2}-\dfrac{3}{n^3}\right)}=0\)
\(a,lim\dfrac{4n^5-3n^2}{\left(3n^2-2\right)\left(1-4n^3\right)}\)
\(=lim\dfrac{4-\dfrac{3}{n^3}}{\left(3-\dfrac{2}{n^2}\right)\left(\dfrac{1}{n^3}-4\right)}\)
\(=\dfrac{4-0}{\left(3-0\right)\left(0-4\right)}=\dfrac{4}{-12}=-\dfrac{1}{3}\)
\(\lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}=\lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(1-\dfrac{10}{n}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(3-\dfrac{3}{n}\right)^3}=\dfrac{1.1^2}{1.3}=\dfrac{1}{3}\)
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) \(lim\left(3n^3+2n^2+n\right)\)
\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)
\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)=+\infty.3=+\infty\)
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/ \(=-n^3\left(2+\frac{3}{n}-\frac{5}{n^2}+\frac{2020}{n^3}\right)=-\infty.2=-\infty\)
CMR: với mọi số tự nhiên n thì:
a)\(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\) chia hết cho 5
b)\(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)chia hết cho 2
a, Ta có: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
\(\Rightarrowđpcm\)
b, \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+31n+5-6n^2-7n+5\)
\(=24n+10=2\left(12n+5\right)⋮2\)
\(\Rightarrowđpcm\)
a)
= n3 + 2n2 + 3n2 + 6n - n - 2 + 2
= 5n2 + 5n
= 5(n2 + n ) chia hết cho 5
b)
= 2(12n +5) chia hết cho 2
Tìm số tự nhiên n thỏa mãn :
\(a,5\left(2-3n+42+3n\right)\ge0\)
\(b, \left(n+1\right)^2-\left(n-2\right)\left(n+2\right)\le1,5\)