1) \(x^4-10x^2+1=0\)
\(\Leftrightarrow\left(x^2-5\right)^2=24\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5=2\sqrt{6}\\x^2-5=-2\sqrt{6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5+2\sqrt{6}}\\x=\sqrt{5-2\sqrt{6}}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\\x=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}+\sqrt{2}\\x=\sqrt{3}-\sqrt{2}\end{matrix}\right.\)
Vậy \(x=\sqrt{2}+\sqrt{3}\) là một nghiệm của pt
2) Ta có: \(a-b=\sqrt{3}+1,b-c=\sqrt{3}-1\)
\(\Rightarrow a-c=a-b+b-c=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)
\(A=a^2+b^2+c^2-ab-bc-ac\)
\(\Rightarrow2A=2a^2+2b^2+2c^2-2ab-2ac-2bc\)
\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)
\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)
\(\left(\sqrt{3}+1\right)^2+\left(\sqrt{3}-1\right)^2+\left(2\sqrt{3}\right)^2\)
\(=3+2\sqrt{3}+1+3-2\sqrt{3}+1+12=20\)
\(\Rightarrow A=10\in N\)
3) \(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}=\sqrt{6+2\sqrt{6}+2\sqrt[]{3}+2\sqrt{2}}=\sqrt{\left(2+2\sqrt{2}+1\right)+2\left(\sqrt{2}+1\right)\sqrt{3}+3}=\sqrt{\left(\sqrt{2}+1\right)^2+2\left(\sqrt{2}+1\right)\sqrt{3}+3}=\sqrt{\left(\sqrt{2}+1+\sqrt{3}\right)^2}=\sqrt{2}+1+\sqrt{3}\)
\(\Rightarrow\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{3}=\sqrt{2}+1\)
