Let f (x) = 1 / x, G (x) = ax & # 178; + BX (a, B ∈ R, a ≠ 0) If the image of y = f (x) and the image of y = g (x) have and only have two different common points a (x1, Y1), B (X2, Y2), then the following judgment is correct x1+x2>0,y1+y2>0 x1+x2>0,y1+y2

Let f (x) = 1 / x, G (x) = ax & # 178; + BX (a, B ∈ R, a ≠ 0) If the image of y = f (x) and the image of y = g (x) have and only have two different common points a (x1, Y1), B (X2, Y2), then the following judgment is correct x1+x2>0,y1+y2>0 x1+x2>0,y1+y2

If the image of y = f (x) and the image of y = g (x) have and only have two different common points a (x1, Y1), B (X2, Y2),
That is, 1 / x = ax & # 178; + BX has and only has two different solutions
That is ax & # 179; + BX & # 178; - 1 = 0, there are only two different solutions
∴ ax³+bx²-1=a(x-x1)²(x-x2)
That is ax & # 179; + BX & # 178; - 1 = a * [x & # 179; - (2x1 + x2) x & # 178; + (x1 & # 178; + 2x1 · x2) x - X1 & # 178; · x2]
∴ x1²+2x1·x2=0,ax1²·x2=1
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