\(\frac{2^4\cdot3^5}{6^4}\)

\(=\frac{2^4\cdot3^5}{2^4\cdot3^4}\)

\(=\frac{2̸^4\cdot3̸^4\cdot3}{2̸^4\cdot3̸^4}\)

\(=3\)

\(\frac{2^4.3^5}{6^4}=\frac{2^4.3^4.3}{6^4}=\frac{\left(2.3\right)^4.3}{6^4}=\frac{6^4.3}{6^4}=3\)

Đặt \(\hept{\begin{cases}\left(x+\frac{1}{x}\right)^3=a\\x^3+\frac{1}{x^3}=b\end{cases}}\)

Ta có

\(A=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+2+\frac{1}{x^6}\right)}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)

\(=\frac{a^2-b^2}{a+b}=a-b\)

\(=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)

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

\(A=\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)

\(=\frac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\frac{6}{\sqrt{3}}\)

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

\(=10-2\sqrt{3}+\frac{6}{\sqrt{3}}\)

\(=\frac{10\sqrt{3}-6+6\sqrt{3}}{\sqrt{3}}\)

\(=\frac{16\sqrt{3}-6}{\sqrt{3}}\)

Mình rút gọn như sau:

   \(\frac{6^3+3.6^2+3^3}{-13}\)

\(=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}\)

\(=\frac{3^2.\left(2^3+3.2^2+3\right)}{-13}\)

\(=\frac{9.\left(8+12+3\right)}{-13}\)

\(=\frac{9.23}{-13}=\frac{207}{-13}\)

Vì tử số và mẫu số ko cùng chia hết cho 1 số nào ngoài 1 suy ra phân số trên tối giản

\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{2\sqrt{3}+\sqrt{18}+2\sqrt{3}-\sqrt{18}}{4-6}\right)-\frac{1}{\sqrt{2}}.\)

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

\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{2\sqrt{6}-6}{\sqrt{2}+1}-\frac{1}{\sqrt{2}}\)