(1a) f(x)=
{ 5-7x for x<5/7
{ 7x-5 for x≥5/7
Note that 5/7=20/28=50/70.
(1b) 7x-5≤3x-2⇒4x≤3, x≤¾, x≤21/28, so 20/28≤x≤21/28, 5/7≤x≤¾.
5-7x≤3x-2⇒7≤10x, x≥7/10, x≥49/70, so 49/70≤x≤50/70, 7/10≤x≤5/7.
Therefore 7/10≤x≤¾.
(2a) 2z₂=-4+8i, 3z₁=3-3i,
2z₂-3z₁=-7+11i,
|-7+11i|²=49+121=170 (assuming || means magnitude).
(2b) z₁-z₂=3-5i, conjugate is 3+5i;
z₁z₂=(1-i)(-2+4i)=-2+6i+4=2+6i, conjugate is 2-6i.
Sum of conjugates: 5-i.
|5-i|³=(√(25+1))³=26√26.
(3) (x-1)[(x-1)(x+3)+3]
+5[4(x+3)-4]
+5[-12-4(x-1)]=
(x-1)(x²+2x)
+5(4x+8)
-5(4x+8)=
x³+x²-2x=0,
x(x+2)(x-1)=0.
Solution: x=0, or -2 or 1.