a) lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)

= lim \(\frac{\left(2-\frac{3}{n}+\frac{5}{n^2}\right)\left(2+\frac{1}{n}\right)}{\left(\frac{4}{n}-3\right)\left(2+\frac{1}{n}+\frac{1}{n^2}\right)}=\frac{4}{-6}=-\frac{2}{3}\)

b)lim ( \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\))

= lim ( \(\frac{n\sqrt{n^4+1}-\sqrt{4n^6+2}}{n^2}\) )

= lim \(\frac{\left(n^6+n^2\right)-\left(4n^6+2\right)}{n^2\left(n\sqrt{n^4+1}+\sqrt{4n^2+2}\right)}\)

= lim \(\frac{-3n^6+n^2+2}{n^3\sqrt{n^4+1}+n^2\sqrt{4n^2+2}}\)

= lim \(\frac{-3n\left(1-\frac{1}{n^4}-\frac{2}{n^6}\right)}{\sqrt{1+\frac{1}{n^4}}+\frac{1}{n^2}\sqrt{4+\frac{2}{n^2}}}\)

= lim \(-3n=-\infty\)

c) lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)

= lim\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)

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: \(=\left(n+1\right)\left(n^2+2n\right)\)

\(=n\left(n+1\right)\left(n+2\right)⋮6\)

a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)

b: 

Ta chứng minh \(2^2+4^2+...+\left(2n\right)^2=\frac{2n\left(n+1\right)\left(2n+1\right)}{3}\)  (1)  

với mọi n \(\in\)N* , bằng phương pháp quy nạp 

Với n = 1, ta có \(2^2=4=\frac{2.1\left(1+1\right)\left(2.1+1\right)}{3}\)

=> (1) đúng khi n = 1 

Giả sử đã có (1) đúng khi n = k , k\(\in\)N* , tức là giả sử đã có : 

\(2^2+4^2+...+\left(2k\right)^2=\frac{2k\left(k+1\right)\left(2k+1\right)}{3}\)

Ta chứng minh (1) đúng khi n = k + 1 , tức là ta sẽ chứng minh 

\(2^2+4^2+...+\left(2k\right)^2+\left(2k+2\right)^2=\frac{2k\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{3}\)

=> Từ giả thiết quy nạp ta có : 

\(2^2+4^2+...+\left(2k\right)^2+\left(2k+2\right)^2=\frac{2k\left(k+1\right)\left(2k+1\right)}{3}+\left(2k+2\right)^2\)

                                                                    \(=\frac{2\left(k+1\right)\left(2k^2+k+6k+6\right)}{3}\)

                                                                    \(=\frac{2\left(k+1\right)\left[2k\left(k+2\right)+3\left(k+2\right)\right]}{3}\)

                                                                    \(=\frac{2\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{3}\)

Từ các chứng minh trên , suy ra (1) đúng với mọi n \(\in\)N*                                             

ai quan tam lam chi

a/ \(lim\left(\sqrt[3]{n-n^3}+n+\sqrt{n^2+3n}-n\right)\)

\(=lim\left(\frac{n}{\sqrt[3]{\left(n-n^3\right)^2}-n\sqrt[3]{\left(n-n^3\right)}+n^2}+\frac{3n}{\sqrt{n^2+3n}+n}\right)\)

\(=lim\left(\frac{1}{\sqrt[3]{n^3+2n+\frac{1}{n}}+\sqrt[3]{n^3-n}+n}+\frac{3}{\sqrt{1+\frac{3}{n}}+1}\right)=0+\frac{3}{1+1}=\frac{3}{2}\)

b/ \(lim\left(\frac{-2\sqrt{n}-4}{\sqrt{n-2\sqrt{n}}+\sqrt{n+4}}\right)=lim\left(\frac{-2-\frac{4}{\sqrt{n}}}{\sqrt{1-\frac{2}{\sqrt{n}}}+\sqrt{1+\frac{4}{n}}}\right)=-\frac{2}{1+1}=-1\)

c/ \(lim\left(\frac{3n^2}{\sqrt[3]{n^6+6n^5+9n^4}+\sqrt[3]{n^6+3n^5}+n^2}\right)=lim\left(\frac{3}{\sqrt[3]{1+\frac{6}{n}+\frac{9}{n^2}}+\sqrt[3]{1+\frac{3}{n}}+1}\right)=\frac{3}{3}=1\)

d/ \(lim\left(\sqrt[3]{n^3+6n}-n+n-\sqrt{n^2-4n}\right)=lim\left(\frac{6n}{\sqrt[3]{n^6+12n^4+36n^2}+\sqrt[3]{n^6+6n^4}+n^2}+\frac{4n}{n+\sqrt{n^2-4n}}\right)\)

\(=lim\left(\frac{6}{\sqrt[3]{n^3+12n+\frac{36}{n}}+\sqrt[3]{n^3+6n}+n}+\frac{4}{1+\sqrt{1-\frac{4}{n}}}\right)=0+\frac{4}{1+1}=2\)

e/ \(lim\left(\frac{-3.3^n+4.4^n}{5.3^n+\frac{3}{2}.4^n}\right)=lim\left(\frac{-3\left(\frac{3}{4}\right)^n+4}{5.\left(\frac{3}{4}\right)^n+\frac{3}{2}}\right)=\frac{0+4}{0+\frac{3}{2}}=\frac{8}{3}\)

f/ \(lim\left(\frac{9^n-5.5^n+7.7^n}{9.3^n+5^n+2.8^n}\right)=lim\left(\frac{1-5.\left(\frac{5}{9}\right)^n+7\left(\frac{7}{9}\right)^n}{9.\left(\frac{1}{3}\right)^n+\left(\frac{5}{9}\right)^n+2.\left(\frac{8}{9}\right)^n}\right)=\frac{1}{0}=+\infty\)

g/ \(lim\left(\frac{6.6^n+3^5.9^n}{3^3.9^n-\frac{1}{2}.4^n}\right)=lim\left(\frac{6\left(\frac{2}{3}\right)^n+3^5}{3^3-\frac{1}{2}\left(\frac{4}{9}\right)^n}\right)=\frac{3^5}{3^3}=9\)

a) (2n-1)⁴ : (2n-1) = 27

(2n-1)³ = 27  =3³

=> 2n - 1= 3

=> 2n = 4

n = 2

phần b,c làm tương tự nha bn

d) (21+n) : 9 = 9⁵:9⁴

(2n+1) : 9 = 9

2n + 1 = 81

2n = 80

n = 40

Tìm số tự nhiên n, biết :

a/ (2.n−1)4:(2.n−1)=27

\(\left(2.n-1\right)^3=27\)

\(2.n-1=3^3\Rightarrow2.n-1=3\)

2.n - 1 = 3

2.n      = 3 + 1

   n      =  4 : 2

   n      =  2

B,C tương tự nha

d) \(\left(21+n\right):9=9^5:9^4\)

   \(\left(21+n\right):9=9\)

        \(21+n=9.9\)

         \(21+n=81\)

                 \(n=81-21\)

                  \(n=60\)

\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)

\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)

\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)

\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)

\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)