Cho [tex]f(x)=\frac{x+\sqrt{5}}{\sqrt{x}+\sqrt{x+\sqrt{5}}}+\frac{x-\sqrt{5}}{\sqrt{x}-\sqrt{x-\sqrt{5}}}[/tex]. Tính f(3)
Bây giờ chứng minh:
Với $a;b>0$ và $a^2-b>0$ thì:
[tex]\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}};\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}[/tex]
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Đặt: [tex]x=\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}\Rightarrow x>0\Rightarrow x=\sqrt{x^2}[/tex]
Xét: [tex]x^2=2a+2\sqrt{a^2-b}=4(\frac{a+\sqrt{a^2-b}}{2})\Rightarrow x=2\sqrt{\frac{a+\sqrt{a^2-b}}{2}}[/tex]
[tex]\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}=2\sqrt{\frac{a+\sqrt{a^2-b}}{2}}[/tex]
CMTT; [tex]\sqrt{a+\sqrt{b}}-\sqrt{a-\sqrt{b}}=2\sqrt{\frac{a-\sqrt{a^2-b}}{2}}[/tex]
[tex]\Rightarrow \sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}[/tex]
[tex]\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}[/tex]
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Áp dụng: [tex]f(x)=\frac{x+\sqrt{5}}{\sqrt{x}+\sqrt{x+\sqrt{5}}}+\frac{x-\sqrt{5}}{\sqrt{x}-\sqrt{x-\sqrt{5}}}\Rightarrow f(x)=\frac{3+\sqrt{5}}{\sqrt{3}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{3}-\sqrt{3-\sqrt{5}}}[/tex]
[tex]f(x)=\frac{3+\sqrt{5}}{\sqrt{3}+\frac{\sqrt{5}+1}{\sqrt{2}}}+\frac{3-\sqrt{5}}{\sqrt{3}-\frac{\sqrt{5}-1}{\sqrt{2}}}=\sqrt{2}(\frac{3+\sqrt{5}}{\sqrt{6}+\sqrt{5}+1}+\frac{3-\sqrt{5}}{\sqrt{6}-\sqrt{5}+1})=\sqrt{2}(4-\sqrt{6})[/tex]
Vậy: [tex]f(x)=\sqrt{2}(4-\sqrt{6})[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
