
\(\sqrt{13+6.\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
b) (
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(a,\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\) \(=\sqrt{13+6\sqrt{4+\sqrt{1-2.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}}\) \(=\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\) \(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\) \(=\sqrt{13+6\sqrt{3+2\sqrt{2}}}\) \(=\sqrt{13+6\sqrt{1+2\sqrt{2}+2}}\) \(=\sqrt{13+6\sqrt{\left(1+\sqrt{2}\right)^2}}\) \(=\sqrt{13+6\left(1+\sqrt{2}\right)}=\sqrt{13+6+\sqrt{12}}\) \(=\sqrt{19+2\sqrt{3}}\) a) = \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\) = \(\sqrt{13+6\sqrt{4+\sqrt{8-2.2\sqrt{2}+1}}}\) = \(\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\) = \(\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\) = \(\sqrt{13+6\sqrt{2+2\sqrt{2}+1}}\) = \(\sqrt{13+6\left(\sqrt{2}+1\right)}\) = \(\sqrt{13+6\sqrt{2}+6}=\sqrt{19+6\sqrt{2}}\) = \(\sqrt{18+2.3\sqrt{2}+1}\) = \(\sqrt{\left(3\sqrt{2}+1\right)^2}\) = \(3\sqrt{2}+1\) a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\) \(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\) Thay x = 4 => \(\sqrt{x}=2\) vào B ta được : \(B=\frac{2+5}{2-3}=-7\) b, Ta có : Với \(x\ge0;x\ne9\) \(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\) \(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\) \(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\) Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
