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

\(\Leftrightarrow\sqrt{1+8}=3\)

\(\Leftrightarrow\sqrt{9}=3\)

mà \(\sqrt{9}=\sqrt{3^2}=\left|3\right|=3\)

\(\Leftrightarrow3=3\)

\(\Rightarrow\sqrt{1^3+2^3}=1+2\)

mấy bài khác chị giải tương tự là ra.

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chứng minh \(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2\)

Chứng minh cái tổng quát:

\(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)

Ta dễ thấy:

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

Từ đó ta có:

\(\left\{{}\begin{matrix}1^3=1^2-0^2\\2^3=\left(1+2\right)^2-1^2\\.........................\\n^3=\left(1+2+...+n\right)^2-\left(1+2+....+\left(n-1\right)\right)^2\end{matrix}\right.\)

Cộng tất cả vế theo vế ta được

\(1^3+2^3+...+n^3=\left(1+2+...+n\right)^2\)

đánh lại đề đi đề lỗi nhiều quá

mẫu rút gọn như sau:

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

xong cộng với cái ở ngoài lại ra 4+2 căn 3 làm tương tự

a) \(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}=\frac{\left(3-\sqrt{2}\right)+\left(3+\sqrt{2}\right)}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\frac{6}{3^2-\left(\sqrt{2}\right)^2}=\frac{6}{7}\)

b) \(\frac{2}{3\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{3}+3\sqrt{3}}=\frac{2\left(2\sqrt{3}+3\sqrt{3}\right)-3\left(3\sqrt{2}-3\sqrt{3}\right)}{\left(3\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{3}+3\sqrt{3}\right)}=\frac{19\sqrt{3}-9\sqrt{2}}{-45+15\sqrt{6}}=-\frac{13\sqrt{3}+10\sqrt{2}}{15}\)c) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{2}=\frac{16}{2}=8\)d) \(\frac{3}{2\sqrt{2}-3\sqrt{3}}-\frac{3}{2\sqrt{2}+3\sqrt{3}}=\frac{3\left(2\sqrt{2}+3\sqrt{3}\right)-3\left(2\sqrt{2}-3\sqrt{3}\right)}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=-\frac{18\sqrt{3}}{19}\)