In the triangle ABC, AB > AC, am is the middle line on the edge of BC, and it is proved that: (AB-AC) / 2

In the triangle ABC, AB > AC, am is the middle line on the edge of BC, and it is proved that: (AB-AC) / 2

Extend am to point D, make MD = ma, connect DC
Then the quadrilateral abdc is a parallelogram
So AC = BD
In the triangle abd,
AB+BD>AD
AB-BD

a. If B is the two real roots of the quadratic equation x & sup2; + nx-1, then B / A + A / b =?

-n^2-2
∵ A and B are two real roots of the quadratic equation x & sup2; + nx-1 with one variable
∴a+b=-n,a·b=-1
b/a+a/b=(a²+b²)/ab=((a+b)²-2ab)/ab

The binary linear equation x-2y / 3 = y = 3x-y + 1 / 2 is reduced to the form ax + by = C

12X-2Y=-3