Решение:
- \( 3a(2a-5) - 4a(a-2) \) = \( 6a^2 - 15a - 4a^2 + 8a \) = \( 2a^2 - 7a \)
- \( b(2b+3) - (b-2)^2 \) = \( 2b^2 + 3b - (b^2 - 4b + 4) \) = \( 2b^2 + 3b - b^2 + 4b - 4 \) = \( b^2 + 7b - 4 \)
- \( (c^2)^{11} : c^{18} \) = \( c^{2 \times 11} : c^{18} \) = \( c^{22} : c^{18} \) = \( c^{22-18} \) = \( c^4 \)
- \(\frac{(n^4)^7
^3}{n^{29}}\)= \(\frac{n^{4 \times 7}
^3}{n^{29}}\)= \(\frac{n^{28}
^3}{n^{29}}\)= \(\frac{n^{28+3}}{n^{29}}\)= \(\frac{n^{31}}{n^{29}}\)= \(n^{31-29}\)= \(n^2\)
Ответ: 1) \(2a^2 - 7a\); 2) \(b^2 + 7b - 4\); 3) \(c^4\); 4) \(n^2\).