16(1 – 2х) – 16 при х = 3/820х – (х – 3) + (10 – 3х) при х = – 1/4–6(2х – 5) – 4х

Решите задачу:

$16(1-2x)-16=16-32x-16=-32x=-32\cdot \frac{3}{8} =-12 \\\ 20x-(x-3)+(10-3x)=20x-x+3+10-3x=16x+13=16\cdot( -\frac{1}{4)}+13= 9 \\\ -6(2x-5)-4x-8=-12x+30-4x-8=-16x+22=-16\cdot \frac{5}{4}= 2 \\\ 2x-(3x-6)-5x=2x-3x+6-5x=6-6x=6-6\cdot0.1=5,4$

16(1 – 2х) – 16 при х = 3/820х – (х – 3) + (10 – 3х) при х = – 1/4–6(2х – 5) – 4х – 8 при...