(р-10)
Ибо
а^2-2ab^2+b^2=(a-b)^2
А (p-10)(p-10)=(p-10)^2=p^2-20p+10^2
1) (х-у)(1-7в)
2) 5у(х+1)+в(х+1)=(х+1)(5у+в)
F(x)=Q₁(x)(4x+10)-14, т.е. F(-5/2)=-14
F(x)=Q₂(x)(9x-3)+37, т.е. F(1/3)=37
F(x)=Q(x)(6x²+13x-5)+(ax+b)=Q(x)(2x+5)(3x-1)+(ax+b), т.е.
F(-5/2)=-5a/2+b=-14 и F(1/3)=a/3+b=37, откуда
17a/6=51, a=18, b=31. Ответ: 18x+31.
Х-1+х+2=20+х -5
х+х-х=20-5+1-2
х=14