Cho $x=ab+\sqrt{(1+a^2)(1+b^2)} ; y=a\sqrt{1+b^2}+b\sqrt{1+c^2}$ ( với ab>0).
Tính y theo x
Cho $x=ab+\sqrt{(1+a^2)(1+b^2)} ; y=a\sqrt{1+b^2}+b\sqrt{1+c^2}$ ( với ab>0).
Started By hoangmanhquan, 03-11-2013 - 19:22
#2
Posted 03-11-2013 - 23:36
Zaraki
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PQT
Cho $x=ab+\sqrt{(1+a^2)(1+b^2)} ; y=a\sqrt{1+b^2}+b\sqrt{1+c^2}$ ( với ab>0).
Tính y theo x
Lời giải. Ta có $$\begin{aligned}y^2 & =a^2+b^2+2a^2b^2+2ab \sqrt{(a^2+1)(b^2+1)} \\ & = (a^2+1)(b^2+1)-1+a^2b^2+2ab \sqrt{(a^2+1)(b^2+1)} \\ & = \left( ab+ \sqrt{(a^2+1)(b^2+1)} \right)^2-1=x^2-1 \end{aligned} $$
Vậy $y= \sqrt{x^2-1}$.
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#3
Posted 05-11-2013 - 18:56
phamphucat
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Hạ sĩ
Bài này nếu lấy $y$ bình phương thì dễ thấy đẳng thức hơn và cũng tự nhiên ít mò hơn
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