Căn bậc hai. Căn bậc ba - Hỏi đáp

\(\left(\frac{15\left(\sqrt{6}-2\right)\left(3-\sqrt{6}\right)+4\left(\sqrt{6}+1\right)\left(3-\sqrt{6}\right)-12\left(\sqrt{6}+1\right)\left(\sqrt{6}-2\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-2\right)\left(3-\sqrt{6}\right)}\right)=\left(\frac{15\left(3\sqrt{6}-6-6+2\sqrt{6}\right)+4\left(3\sqrt{6}-6+3-\sqrt{6}\right)-12\left(6-2\sqrt{6}+\sqrt{6}-2\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-2\right)\left(3-\sqrt{6}\right)}\right)=(\frac{\left(15\left(5\sqrt{6}-12\right)+4\left(2\sqrt{6}-3\right)-12\left(4-\sqrt{6}\right)\right)}{18-7\sqrt{6}})=\left(\frac{75\sqrt{6}-180+8\sqrt{6}-12-48+12\sqrt{6}}{18-7\sqrt{6}}\right)=\left(\frac{95\sqrt{6}-240}{18-7\sqrt{6}}\right)=\left(\frac{95\sqrt{6}-240}{7\sqrt{6}-18}\right)\)

\(\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{\sqrt{2}^2+\sqrt{5}^2+1^2+2\sqrt{2}+2\sqrt{5}+2\sqrt{2}.\sqrt{5}}-\sqrt{\sqrt{5}^2+2\sqrt{10}+\sqrt{2}^2}\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

\(=\sqrt{5}+\sqrt{2}+1-\sqrt{5}-\sqrt{2}\)

\(=1\)

\(M=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\\ M^2=4+\sqrt{7}+4-\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}\\ M^2=8-2\sqrt{16-7}=8-2\sqrt{9}=2\\ \Rightarrow M=\sqrt{2}\)

\(M=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{\left(\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\dfrac{\sqrt{7}}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{\sqrt{7}}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}=\sqrt{2}\)

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