Mixed operation problem of rational number addition and subtraction The more, the better

Mixed operation problem of rational number addition and subtraction The more, the better

-5+21*8/2-6-59
68/21-8-11*8+61
-2/9-7/9-56
4.6-(-3/4+1.6-4-3/4)
1/2+3+5/6-7/12
[2/3-4-1/4*(-0.4)]/1/3+2
22+(-4)+(-2)+4*3
-2*8-8*1/2+8/1/8
(2/3+1/2)/(-1/12)*(-12)
(-28)/(-6+4)+(-1)
2/(-2)+0/7-(-8)*(-2)
(1/4-5/6+1/3+2/3)/1/2
18-6/(-3)*(-2)
(5+3/8*8/30/(-2)-3
(-84)/2*(-3)/(-6)
1/2*(-4/15)/2/3
-3x+2y-5x-7y
Happy every day!

Ask linear equation problem master solution
If the line ax + 2BY-2 = 0 (A and B are both greater than 0) always passes through the point (2,1), then the minimum value of 1 / A + 2 / B is 0

I asked about the equation of degree 3
a+b+c=37
1/4 a-1/5 b=1
1/4 a+b=1/3 c

1/4a=1+1/5b
a=4+4/5b
C = (1 / 4A + b) times 3
c=3+18/5b
4+4/5b+b+3+18/5b=37
b=50/9
1 / 4a-1 / 5 times 50 / 9 = 1
a=76/9
c=37-76/9-50/9
c=23
a=76/9
b=50/9
c=23

Given the quadratic equation of one variable: (1) mx2-4x + 4 = 0; (2) x2-4mx + 4m2-4m-5 = 0 (m ∈ z), it is necessary and sufficient to find that the roots of equations (1) and (2) are integers

Equation (1) has real root ⇔ △ 1 = 16-16m ≥ 0, that is, m ≤ 1, and m ≠ 0. Equation (2) has real root ⇔ △ 2 = 16m2-4 (4m2 − 4m − 5) ≥ 0 {m ≥ - 54, and m ≠ 0. From − 54 ≤ M ≤ 1 and m ∈ Z, M = - 1, 1. When m = - 1, equation (1) is x2 + 4x-4 = 0, and there is no integer solution; when m = 1, equation (1) has integer solution x = 2, and equation (2) has integer solution x = - 1 or 5, thus (1), (2) and (2) have integer solution x = - 1 On the other hand, from M = 1, we can deduce that equations (1) and (2) have integer solutions, so the necessary and sufficient condition for equations (1) and (2) to have integer solutions is m = 1