第27页 - 苏科版数学补充习题七年级上下册答案

$ ={({a}^{1+2+m})}^2$

$ ={a}^{2(3+m)}$

$ ={a}^{6+2m}$

$ =(-\frac 12)^5x^{2×5}y^5$

$ =-\frac 1{32}x^{10}y^5$

$ ={-a}^6\cdot {a}^{10}$

$ ={-a}^{16}$

$ ={({a}^3{b}^6)}^4$

$ ={a}^{12}{b}^{24}$

$ ={x}^8+{x}^8-9{x}^8$

$ =-7{x}^8$

$ ={(x+2y)}^8\cdot {(x+2y)}^2$

$ ={(x+2y)}^{10}$

$ y^4$

$ -\frac 12a^3b$

$ ab$

$ a^m$

$ ={(-\frac {1} {8})}^5×4^{10}$

$ ={(-2^{-3})}^5×{(2^2)}^{10}$

$ =-2^{-15}×2^{20}$

$ ={-2}^5$

$ =-32$

$ ={(\frac {7} {3})}^{2000}×{(\frac {3} {7})}^{2000}×{(\frac {3} {7})}^2$

$ ={(\frac {7} {3}×\frac {3} {7})}^{2000}×{(\frac {3} {7})}^2$

$ =\frac {9} {49}$