Find the zero point of function y = (3x ^ 2-x ^ 2) / (x ^ 2-1) + 1 / (1-x) - 2 | Math enthusiast | Mathematical art

Find the zero point of function y = (3x ^ 2-x ^ 2) / (x ^ 2-1) + 1 / (1-x) - 2

Let f (x) = - 1 / 3x ^ 3 + x ^ 2 + (m ^ 2-1) x, where m is greater than 0. Given that f (x) has three mutually unequal zeros 0, x1, X2, and 10 - is 14 days and 23 hours from the end of the problem, X1 is less than x2. If for any x belonging to [x1, X2], f (x) is greater than f (1), the value range of M is 09

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9. For any x1, X2 (x1 ≠ x2) in the domain of definition of function f (x), we have the following conclusion: (1) f (x1 + x2) = f (x1) * f (x2) (2) f (x1 * x2) = f (x1) + F (x) (1)f(x1+x2)=f(x1)*f(x2) (2)f(x1*x2)=f(x1)+f(x2) (3)[f(x1)-f(x2)]/(x1-x2)>0 (4) f[(x1+x2)/2]
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11. The number of zeros of function y = x ^ 2-3x + 1
12. Finding the zeros of the function y = (3x-x ^ 2) / (x ^ 2-1) + 1 / (1-x) - 2
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