计算:(1)(J)?(AB2+3A2B%1);(2)(34x6y...

(1)原式=4a2b4(3a2b-2ab-1)=12a4b5-8a3b5-4a2b4;(2)原式=9a2-18ab+9b2-(9a2-b2)=9a2-18ab+9b2-9a2+b2=8b2-18ab.

(1)aa2a3=a6;(2)(a4)2(-a2)3=-a8*a6=-a14 ;(3)(-3a2bc)2(-2ab2)3 =9a4b2c2*(-8)a3b6=-72a7b8c2;(4)23a7b5÷32a2b5=49a5;(5)xm+n÷xn=xm;(6)(-2a2b)2(ab2-a2b+a3)=4a5b4-4a

(1)(3a2b)2+(8a6b3)÷(-2a2b),=9a4b2-(8÷2)*(a6-2b3+1),=9a4b2-4a4b2,=5a4b2;(2)(a+3b-4)(a-3b-4),=(a-4)2-9b2,=a2-8a+16-9b2.

(1)原式=1 3x2 ;(2)原式=ab2 2c2 ?4cd ?3a2b2 =?2d 3ac ;(3)原式=24xy 7z ?1 ?8xyz =?3 7z2 ;(4)原式=a6b3 c6 ??c2 a2b ?a4 b4c4 =?a8 b2c8 .

(4a3b-10b3)+(-3a2b2+10b3)=4a3b-10b3-3a2b2+10b3=4a3b-3a2b2(4x2y-5xy2)-(3x2y-4xy2)=4x2y-5xy2-3x2y+4xy2=x2y-xy2

小手一抖金币到手

#include <stdio.h> int main() { int sum = 0; for (int i = 1; i < 21 ; i++) sum += i * i * ( i % 2 ? 1 : -1); //条件运算符 + 非零即真,不懂就百度吧. printf("%d\n", sum); return 0; }

(1)原式=-12a4b2+6a3b3-3a3b2;(2)原式=-5x3y+5x2y2-x3y-2x2y2=-5x3y+3x2y2-x3y.

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