Question

[Khuất Ngọc Nam]

Giúp em phần II ạ. Được bao nhiêu cx đc. Em xin cảm ơn

Answer 1

[Lấp La Lấp Lánh]

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

Answer 2

[Lấp La Lấp Lánh]

2) Ta có: \(a-b=\sqrt{3}+1,b-c=\sqrt{3}-1\)

\(\Rightarrow a-c=a-b+b-c=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)

\(A=a^2+b^2+c^2-ab-bc-ac\)

\(\Rightarrow2A=2a^2+2b^2+2c^2-2ab-2ac-2bc\)

\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)

\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)

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

\(=3+2\sqrt{3}+1+3-2\sqrt{3}+1+12=20\)

\(\Rightarrow A=10\in N\)