\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)

\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)

\(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)

\(=\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}\cdot1+1}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)

\(=\left(\sqrt{3}-1\right)^2\left(2+\sqrt{3}\right)\)

\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)

\(=8+4\sqrt{3}-4\sqrt{3}-6\)

\(=2\)

\(C=-\left[\dfrac{1}{3}\cdot\dfrac{\left(3+1\right)\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{\left(4+1\right)\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{\left(50+1\right)\cdot50}{2}\right]\\ C=-\left(\dfrac{1}{3}\cdot\dfrac{4\cdot3}{2}+\dfrac{1}{4}\cdot\dfrac{5\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{51\cdot50}{2}\right)\\ C=-\left(2+\dfrac{5}{2}+...+\dfrac{51}{2}\right)\\ C=-\dfrac{4+5+...+51}{2}=-\dfrac{\dfrac{\left(51+4\right)\left(51-4+1\right)}{2}}{2}=-\dfrac{55\cdot48}{4}=-660\)

Chắc có lẽ bạn định làm như này:

\(\frac{1}{\left(3n+2\right)\left(3n+5\right)}=\frac{3}{3\left(3n+2\right)\left(3n+5\right)}=\frac{\left(3n+5\right)-\left(3n+2\right)}{3\left(3n+2\right)\left(3n+5\right)}=\frac{1}{3}\left[\frac{1}{3n+2}-\frac{1}{3n+5}\right]\)

n=1=> đẳng thức đúng

giả sử có số n=a thoả mãn pt=>

2+5+8+....+(3a-1)=a(3a+1)/2=(3a^2+a)/2(1)

phải chứng minh n=a+1 thoả mãn pt:

2+5+8+......+(3a+2)=(a+1)(3a+4)/2=(3a^2+7a+4)/2(2)

lấy (2) trừ (1) ta được:

(6a+4)/2=3a+2

=> 0=0 (đúng vs mọi a)

=> đẳng thức (2) đúg, dpcm

Gọi ĐTV hay lê chí cường ấy

Đặt A = 2 + 5+  ....... + (2n - 1)

Số các số hạng là:

(3n - 1 - 2)/3 + 1 = (3n - 3)/3 + 1 = n - 1 + 1 = n

A = n x (3n -1 + 2) : 2

A = \(\frac{n\left(3n+1\right)}{2}\) => DPCM 

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

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

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

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

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

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

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

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

.........