X^2(x^2-3x+1)-2x(x^3-3x^2+x)+x^4-3x^3+x^3 при x=1 1/3

Question

X^2(x^2-3x+1)-2x(x^3-3x^2+x)+x^4-3x^3+x^3 при x=1 1/3

Comments

а в последнем не квадрат?

---

т.е. степень вторая?

Answer 1

X²(x²-3x+1)-2x(x³-3x²+x)+x⁴-3x³+x³=x⁴-3x³+x²-2x⁴+6x³-2x²+x⁴-3x³+x³=
=x³-x²
x³-x²
x=1 1/3=4/3
(4/3)³-(4/3)²=(64/27)-(16/9)=64/27-48/27=16/27