[Toán 12] sao cho diện tích tam giác OAB = 1/4

B

behang_95

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N

nguyenbahiep1

[TEX]y=\frac{2x}{x+1} \\ txd : x \not= -1 \\ M ( x_0, \frac{2x_0}{x_0+1} \\ pttt :(d): y = \frac{2}{(x_0+1)^2} .(x-x_0) + \frac{2x_0}{x_0+1} \\ (d) cut ox \\ 0 = \frac{2}{(x_0+1)^2} .(x-x_0) + \frac{2x_0}{x_0+1} \\ \Rightarrow A ( -x_0^2 , 0 ) \Rightarrow |\vec{OA}| = x_0^2\\ (d) cut oy \\ y = \frac{2}{(x_0+1)^2} .(0-x_0) + \frac{2x_0}{x_0+1} \\ \Rightarrow B (0, \frac{2}{(x_0+1)^2}) \Rightarrow |\vec{OB}| = \frac{2}{(x_0+1)^2}\\ S_{OAB} = \frac{1}{2}.\frac{2.x_0^2}{(x_0+1)^2} = \frac{1}{4} \\ x_0 = 1 \Rightarrow M ( 1,1) \\ x_0 = - \frac{1}{3} \Rightarrow M (- \frac{1}{3}, -1) [/TEX]
 
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