Решение:
- a) \( 2\frac{2}{3} = \frac{2 · 3 + 2}{3} = \frac{8}{3} \).
- \( \frac{8}{3} x^2y^3 \left(-\frac{1}{2}xy^2\right)^4 = \frac{8}{3} x^2y^3 \left(-\frac{1}{2}\right)^4 x^4 (y^2)^4 \)
- \( = \frac{8}{3} x^2y^3 · \frac{1}{16} x^4 y^8 \)
- \( = \left(\frac{8}{3} · \frac{1}{16}\right) (x^2 · x^4) (y^3 · y^8) \)
- \( = \frac{8}{48} x^{2+4} y^{3+8} = \frac{1}{6} x^6 y^{11} \)
- б) \( x^{a-2} · x^{3-a} · x^1 = x^{(a-2) + (3-a) + 1} \)
- \( = x^{a-2+3-a+1} = x^{(a-a) + (-2+3+1)} = x^{0+2} = x^2 \)
Ответ: a) \( \frac{1}{6} x^6 y^{11} \); б) \( x^2 \).