A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)

Xét: n⁴ + 4 = (n²+2)² - 4n² = (n²-2n+2)(n²+2n+2) = [(n-1)²+1][(x+1)²+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)

B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)

Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)

= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)

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

= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)

= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)

Vậy ...

c: \(\left(n-2\right)^2-\left(n+3\right)\left(n-3\right)=4\left(n-1\right)\)

\(\Leftrightarrow n^2-4n+4-n^2+9=4n-4\)

=>-4n+13=4n-4

=>-8n=-17

hay n=17/8

a: \(\left(n-2\right)\left(n+2\right)+6\left(n-1\right)=\left(n+1\right)^2\)

\(\Leftrightarrow n^2-4+6n-6=n^2+2n+1\)

=>6n-10=2n+1

=>4n=11

hay n=11/4

d: \(2\left(3-x\right)-3\left(x-1\right)=4\left(x-3\right)\)

=>6-2x-3x+3=4x-12

=>-5x+9=4x-12

=>-9x=-21

hay x=7/3

\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)

\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)=+\infty.3=+\infty\)

\(Q=n^3+\left(n+1\right)^3+\left(n+2\right)^3⋮9\)

\(Q=n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8\)

\(Q=3n^3+9n^2+15n+9\)

\(Q=3n\left(n^2+5\right)+9\left(n^2+1\right)\)

mà \(\left\{{}\begin{matrix}9\left(n^2+1\right)⋮9\\3n⋮3\\n^2+5⋮3\end{matrix}\right.\left(\forall n\inℕ^∗\right)\)

\(\Rightarrow Q=3n\left(n^2+5\right)+9\left(n^2+1\right)⋮9,\forall n\inℕ^∗\)

\(\Rightarrow dpcm\)

Vì \(n\in Z^+\)nên\(n\left(n+1\right)\left(n+2\right)>n^3\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)}>n\)

\(\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}>n\)(1)

Lại có:\(n^2+2n+1>n^2+2n\Rightarrow\left(n+1\right)^2>n\left(n+2\right)\Rightarrow\left(n+1\right)^3>n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)}\\ \Rightarrow\sqrt[3]{n^3+3n^2+3n+1}>\sqrt[3]{n^3+3n^2+2n}\)

\(\Rightarrow\sqrt[3]{n^3+3n^2+2n+n+1}>\sqrt[3]{n^3+3n^2+2n+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)

\(\Rightarrow\sqrt[3]{\left(n+1\right)^3}>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)

Tương tự \(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)(2)

Từ (1) và (2) suy ra:

\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< n+1\)

\(n\in Z^+\)nên n² < n² + 2n < n² + 2n + 1 <=> n² < n(n + 2) < (n + 1)² => n³ < n(n + 1)(n + 2) < (n + 1)³ 

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< n+1\)

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n+1}\)\(=\sqrt[3]{\left(n+1\right)\left(n^2+2n+1\right)}=\sqrt[3]{\left(n+1\right)\left(n+1\right)^2}=n+1\)

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)

\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}}< n+1\)

Tiếp tục như vậy,ta có đpcm.

Sorry ! n² < n(n + 2) nên n³ < n(n + 1)(n + 2) (vì n < n + 1)

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) \(\left(21+n\right):9=9^5:9^4\)

   \(\left(21+n\right):9=9\)

        \(21+n=9.9\)

         \(21+n=81\)

                 \(n=81-21\)

                  \(n=60\)