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On the equation of X, does the square of X + 2aX + a = 4 have two unequal real roots? Why?
There must be a solution, which is the discriminant: (2a) squared-4 (a squared-4) > 0
Then 4A square - 4A square + 16 > 0, that is, 16 > 0
So there are two unequal real solutions
X1=2-a X2=-a-2
1/1x2+1/2x3+1/3x4+1/4x5
=4/5
Let's divide first and then calculate
Find the law 1 1 / 8 1 / 27
1.1/8,1/27,1/64,1/125.1/n^3
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