\(A=\left(n^2+2n+1+4\right)^3-\left(n+1\right)^2+2018\)

\(A=\left(\left(n+1\right)^2+4\right)^3-\left(n+1\right)^2+2018\)

ĐẶT:   \(\left(n+1\right)^2=a\)

=>   \(A=\left(a+4\right)^3-a+2018\)

=>   \(A=a^3+12a^2+48a+64-a+2018\)

=>   \(A=\left(a^3-a\right)+12a^2+48a+2082\)

CÓ:

   \(a^3-a=a\left(a-1\right)\left(a+1\right)\)      hiển nhiên chia hết cho 3 và 2 do đây là tích 3 số nguyên liên tiếp 

=>   \(a^3-a⋮6\)

MÀ HIỂN NHIÊN:   \(12a^2+48a+2082⋮6\)

=>    \(A⋮6\)

VẬY TA CÓ ĐPCM.

\(\frac{3}{\sqrt{5}+2}+\frac{1}{\sqrt{2}-1}-\frac{4}{3-\sqrt{5}}\)

\(=\frac{3\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+\frac{\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(=\frac{3\sqrt{5}-6}{5-4}+\frac{\text{​​}\sqrt{2}+1}{2-1}-\frac{4\left(3+\sqrt{5}\right)}{9-5}\)

\(=3\sqrt{5}-6+\text{​​}\sqrt{2}+1-3+\sqrt{5}\)

\(=2\sqrt{5}-8+\text{​​}\sqrt{2}\)

+) Ta chứng minh: \(\frac{x-2}{x+1}\le\frac{x-2}{3}\)

\(\Leftrightarrow\frac{3\left(x-2\right)-\left(x-2\right)\left(x+1\right)}{3\left(x+1\right)}\le0\)'

\(\Leftrightarrow\frac{-\left(x-2\right)^2}{3\left(x+1\right)}\le0\)(luôn đúng)

+) \(6=3\sqrt[3]{xyz}\le x+y+z\)

+) \(\text{Σ}\frac{x-2}{x+1}\le\frac{x-2+y-2+z-2}{3}\le\frac{0}{3}=0\)

Dấu = xảy ra khi x = y = z = 2