Bài 2: 

A = (a+b)(1/a+1/b)

Có: \(a+b\ge2\sqrt{ab}\)

\(\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{ab}}\)

=> \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge2\sqrt{ab}.2\sqrt{\frac{1}{ab}}=4\)

=> ĐPCM

1.b)

Pt (1) : 4(n + 1) + 3n - 6 < 19
<=> 4n + 4 + 3n - 6 < 19 
<=> 7n - 2 < 19
<=> 7n - 2 - 19 < 0
<=> 7n - 21 < 0
<=> n < 3
Pt (2) : (n - 3)^2 - (n + 4)(n - 4) ≤ 43
<=> n^2 - 6n + 9 - n^2 + 16 ≤ 43
<=> -6n + 25 ≤ 43
<=> -6n ≤ 18
<=> n ≥ -3
Vì n < 3 và n ≥ -3 => -3 ≤ n ≤ 3.
Vậy S = {x ∈ R ; -3 ≤ n ≤ 3}

\(b,lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}\)

\(=lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(\dfrac{1}{n}-\dfrac{10}{n^2}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(\dfrac{3}{n^2}-\dfrac{3}{n^3}\right)}=0\)

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

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

\(=\dfrac{4-0}{\left(3-0\right)\left(0-4\right)}=\dfrac{4}{-12}=-\dfrac{1}{3}\)

\(\lim\dfrac{\left(n^2+1\right)\left(n-10\right)^2}{\left(n+1\right)\left(3n-3\right)^3}=\lim\dfrac{\left(1+\dfrac{1}{n^2}\right)\left(1-\dfrac{10}{n}\right)^2}{\left(1+\dfrac{1}{n}\right)\left(3-\dfrac{3}{n}\right)^3}=\dfrac{1.1^2}{1.3}=\dfrac{1}{3}\)

Không ai bt làm::(

\(\lim\dfrac{\left(2n-1\right)\left(3n^2+2\right)^3}{-2n^5+4n^3-1}=\lim\dfrac{\left(\dfrac{2n-1}{n}\right)\left(\dfrac{3n^2+2}{n^2}\right)^3}{\dfrac{-2n^5+4n^3-1}{n^7}}\)

\(=\lim\dfrac{\left(2-\dfrac{1}{n}\right)\left(3+\dfrac{2}{n^2}\right)^3}{-\dfrac{2}{n^2}+\dfrac{4}{n^4}-\dfrac{1}{n^7}}=-\infty\)

\(\lim3^n\left(6.\left(\dfrac{2}{3}\right)^n-5+\dfrac{7n}{3^n}\right)=+\infty.\left(-5\right)=-\infty\)

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

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

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

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

a) (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) \(\left(21+n\right):9=9^5:9^4\)

   \(\left(21+n\right):9=9\)

        \(21+n=9.9\)

         \(21+n=81\)

                 \(n=81-21\)

                  \(n=60\)