回答:原式=-4a^2b-5ab+5ab^2
A^2-4AB-5B^2=(A-5B)(A+B)A^2-4AB-5B^2=A^2-4AB+4B^2-9B^2=(A-2B)^2-9B^2=(A-2B-3B)(A-2B+3B)=(A-5B)(A+B)
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-2b2-5ab-12a2b-5ac-(3)先化简再求值-2y3+.其中|2x-2|+有理数a.b.c在数轴上的对应点如图所示.化简代数式:|2a-b|+3|a+b|-|4c-a|
b-3ab 2 )-(-a 2 b+2ab 2 ) =4a 2 b-3ab 2 +a 2 b-2ab 2 =5a 2 b-5ab 2 则(4a 2 b-3ab 2 )-(-a 2 b+2ab 2 )去括号得 4a 2 b-3ab 2 +a 2 b-2ab 2 ,合并同类项得 5a 2 b-5ab 2 . 故答案是:4a 2 b-3ab 2 +a 2 b-2ab 2 ...
a^2-a b-2b^2+2a+5b-3=(a-2b)(a+b) +2a+5b-3=(a-2b)(a+b)+ 2b -a + 3a+3b-3 =(a-2b)(a+b-1) + 3(a+b-1)=(a-2b+3) (a+b-1)
(ab 2 ) 2 •(-a 3 b) 3 •(-5ab)=a 2 •(b 2 ) 2 •(-1) 3 •(a 3 ) 3 •b 3 •(-5ab)=-a 2 •b 4 •a 9 •b 3 •(-5ab)=a 11 b 7 •5ab =5a 12 b 8 .
(4a^2-5ab+b^2)-(2a^2-3ab+3b^2)=2a^2+8ab-2b^2=2(a^2-b^2+4ab)因为 a^2-b^2=5ab 代入式子中可得2(5ab-4ab)=2ab
解:原式=6a 2 b-10ab 2 -3a 2 b+9ab 2 =3a 2 b-ab 2 把a= ,b=-2代入得原式=- 。
5ab-[2(2a^2-ab-b^2)-3(a^2+b^2)]-3(ab-2b^2)=5ab-[4a²-2ab-2b²-3a²-3b²]-3ab+6b²=2ab+6b²-(a²-2ab-5b²)=2ab+6b²-a²+2ab+5b²=4ab+11b²-a²...