The ellipse X & # 178 / M + Y & # 178 / M-1 = 1 (M > 1) intersects with the straight line y = X-1 at two points a and B. If AB is the left focus F of the ellipse, the value of the real number m is obtained,

The ellipse X & # 178 / M + Y & # 178 / M-1 = 1 (M > 1) intersects with the straight line y = X-1 at two points a and B. If AB is the left focus F of the ellipse, the value of the real number m is obtained,

It is known that the left focus of the ellipse is: (- 1,0)
Let a (x1, Y1)
B（x2,y2）
Because: AF ⊥ BF
y1/(x1+1) * y2/(x2+1) =-1
That is, (x1 + 1) * (x2 + 1) + Y1 * y2 = 0
Because: Y1 = x1-1
y2=x2-1
That is, (x1 + 1) * (x2 + 1) + (x1-1) * (x2-1) = 0
So x1x2 = - 1
Substituting y = X-1 into elliptic equation: (m-1) * x ^ 2 + m (x-1) ^ 2-m ^ 2 + M = 0
That is x ^ 2 (2m-1) - 2mx-m ^ 2 + 2m = 0
Because x1, X2 are two parts of the equation
So x1x2 = (- m ^ 2 + 2m)) / (2m-1)
=-1
M = 2 + √ 3 or 2 - √ 3
Because m > 1
So m = 2 + √ 3

Given that a is a real number, the solution set interval of inequality 2x & # 178; - 12x + a ≤ 0 [1,5]. Find the solution set of inequality | AX-1 | 9

Interval of solution set of inequality 2x & # 178; - 12x + a ≤ 0 [1,5]
The two roots of the equation 2x & # 178; - 12x + a = 0 are 1 and 5
Using Veda's theorem
Then 1 * 5 = A / 2
∴ a=10
∴ |ax-1|≤9
That is, | 10x-1 | ≤ 9
That is - 9 ≤ 10x-1 ≤ 9
That is - 8 ≤ 10x ≤ 10
∴ -4/5≤x≤1
That is to say, the solution set of inequality is [- 4 / 5,1]

If the product of two rational numbers is positive, then the sign of the two factors must be positive______ ．

∵ the product of two rational numbers is positive, and the sign of the two factors must be the same

If the product of two rational numbers is positive, then the sign of the two factors must be positive______ ．

∵ the product of two rational numbers is positive, and the sign of the two factors must be the same