а)
х²+12х+32 = 0
(х+8)(х+4) = 0
х = -8 или х = -4

б)
х²-7х-120 = 0
(х-15)(х+8) = 0
х = 15 or х = -8

х³-х+60 = |х+5|-3
х³-х+60 = х+5-3 or х³-х+60 = -(х+5)-3
х³-х+60 = х+2 or х³-х+60 = -х-8
х³-х-х+8-60 = 0
х³-2х-52 = 0

а)
х²+6х+5 = (х+1)(х+5)
6х²-7х-3 = (3х+1)(2х-3)
5х²+6х+1 = (5х+1)(х+1)
2х²-х-3 = (2х+3)(х-1)

a)

2х-3y=8
7х-5y=-5

Multiplying the first equation by 7 and the second equation by 2, we get:

14х - 21у = 56
14х - 10y = -10

Subtracting the second equation from the first, we get:

-11y = 66
y = -6

Using y = -6 in the first equation:

2x - 3(-6) = 8
2x + 18 = 8
2x = -10
x = -5

So the solution is x = -5, y = -6.

|x-5| = x-5
If x > 5, then x-5 = x-5
If x < 5, then -(x-5) = x-5
So we need to solve the equation for x > 5 and x < 5.

If x > 5:
2x-3 = x-5
x = 2

If x < 5:
-(2x-3) = x-5
-2x+3 = x-5
-3x = -8
x = 8/3

Therefore, the equation has two solutions: x = 2 and x = 8/3.