Question

[Nguyễn Hoàng Vũ]

1. Rút gọn \(A=\frac{\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}}{\sqrt{\left(\sqrt{5}+1\right)\cdot\sqrt{6-2\sqrt{5}}}}\)

2.Tính a) \(B=\left(\sqrt[3]{2}+1\right)^3\cdot\left(\sqrt[3]{2}-1\right)^3\)

b)Tìm C=\(a^3b-ab^3\) với \(a=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\); \(b=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)

3. Giải \(\left|x^2-x+1\right|-\left|x-2\right|=6\)

Answer 1

[Akai Haruma]

Bài 1:
Xét tử số:

\(\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}=\sqrt{3^2+5+2.3\sqrt{5}}-\sqrt{3^2+5-2.3\sqrt{5}}\)

\(=\sqrt{(3+\sqrt{5})^2}-\sqrt{(3-\sqrt{5})^2}=3+\sqrt{5}-(3-\sqrt{5})=2\sqrt{5}\)

Xét mẫu số:
\(\sqrt{(\sqrt{5}+1)\sqrt{6-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{5+1-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{(\sqrt{5}-1)^2}}\)

\(=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}=\sqrt{4}=2\)

Do đó: $A=\frac{2\sqrt{5}}{2}=\sqrt{5}$

Answer 2

[Nguyễn Hoàng Vũ]

dạ em cảm ơn