Ta có: \(\sqrt{n^2 + 1}-\sqrt[3]{3n^3 + 2}\)

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

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

Ta có: \(\sqrt[4]{2n^4 + n + 2}-n\)

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

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

Ta có: \(D = \lim_{n \to +\infty} \frac{\sqrt{n^2 + 1} - \sqrt[3]{3n^3 + 2}}{\sqrt[4]{2n^4 + n + 2} - n}\)

\(=\lim_{n\to+\infty}\frac{n \left( \sqrt{1 + \frac{1}{n^2}} - \sqrt[3]{3 + \frac{2}{n^3}} \right)}{n \left( \sqrt[4]{2 + \frac{1}{n^3} + \frac{2}{n^4}} - 1 \right)}\)

\(=\lim_{n\to+\infty}\frac{\sqrt{1 + \frac{1}{n^2}} - \sqrt[3]{3 + \frac{2}{n^3}}}{\sqrt[4]{2 + \frac{1}{n^3} + \frac{2}{n^4}} - 1}\)

\(=\frac{\sqrt{1 + 0} - \sqrt[3]{3 + 0}}{\sqrt[4]{2 + 0 + 0} - 1}=\frac{1 - \sqrt[3]{3}}{\sqrt[4]{2} - 1}\)

\(F=\lim\limits\dfrac{\sqrt[4]{n^4-2n+1}+2n}{\sqrt[3]{3n^3+n}-n}=\lim\limits\dfrac{\sqrt[4]{\dfrac{n^4}{n^4}-\dfrac{2n}{n^4}+\dfrac{1}{n^4}}+\dfrac{2n}{n}}{\sqrt[3]{\dfrac{3n^3}{n^3}+\dfrac{n}{n^3}}-\dfrac{n}{n}}=\dfrac{1+2}{3-1}=\dfrac{3}{2}\)

Bài giải:

a)1³+2³

=1+8

=9

Vậy 1³+2³ là số chính phương.

b)1³+2³+3³

=1+8+27

=36

Vậy 1³+2³+3³ là số chính phương.

c)1³+2³+3³+4³

=1+8+27+64

=100

Vậy 1³+2³+3³+4³ là số chính phương.

Thank you and study well!

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