Question

[Nguyễn Duy Khang]

Giải chi tiết giúp tui nha, tui cảm ơn nhiều lắm đó, tui thấy khó quá trời luôn mong các bạn giúp đỡ

Tính:

\(a,\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(b,\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}-\sqrt{6+2\sqrt{2}}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

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

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

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

\(=\frac{2+\sqrt{3}+1}{2-\sqrt{3}+1}\)(Vì \(\sqrt{3}>1>0\))

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

Answer 2

[Nguyễn Việt Lâm]

\(a=\frac{2+\sqrt{4+2\sqrt{3}}}{2-\sqrt{4-2\sqrt{3}}}=\frac{2+\sqrt{\left(\sqrt{3}+1\right)^2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{2+\sqrt{3}+1}{2-\sqrt{3}+1}=\frac{3+\sqrt{3}}{3-\sqrt{3}}=\frac{\left(3+\sqrt{3}\right)^2}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)

\(=\frac{12+6\sqrt{3}}{6}=2+\sqrt{3}\)

Xét \(A=\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}>0\)

\(A^2=6+2\sqrt{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}=6+2\sqrt{2}\)

\(\Rightarrow A=\sqrt{6+2\sqrt{2}}\)

\(\Rightarrow\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}-\sqrt{6+2\sqrt{2}}=\sqrt{6+2\sqrt{2}}-\sqrt{6+2\sqrt{2}}=0\)