早教吧作业答案频道 -->数学-->
因式分解a^2b^3-4a^2b,(x+z)^2-(2x-3)^2,9(x-y)^2-6(x-y)+1,16x^4y^4-1,-(m-n)^2+1,4(x+y+z)^2-9(x-y-z)^2(a-3)^2-(2a-6)
题目详情
因式分解
a^2b^3-4a^2b,(x+z)^2-(2x-3)^2,9(x-y)^2-6(x-y)+1,16x^4y^4-1,
-(m-n)^2+1,4(x+y+z)^2-9(x-y-z)^2
(a-3)^2-(2a-6)
a^2b^3-4a^2b,(x+z)^2-(2x-3)^2,9(x-y)^2-6(x-y)+1,16x^4y^4-1,
-(m-n)^2+1,4(x+y+z)^2-9(x-y-z)^2
(a-3)^2-(2a-6)
▼优质解答
答案和解析
1.a^2b^3-4a^2b=a^2b(b^2-4)=a^2b(b+2)(b-2)
2.(x+z)^2-(2x-3)^2=(x+z+2x-3)(x+z-2x+3)=(3x+z-3)(z-x+3)
3.令x-y=a,则原式为9a^2-6a+1=(3a-1)^2=(3x-3y-1)^2
4.16x^4y^4-1=(4^2y^2+1)(4^2y^2-1)=(4^2y^2+1)(2xy+1)(2xy-1)
5.-(m-n)^2+1=1-(m-n)^2=(1+m-n)(1-m+n)
6.4(x+y+z)^2-9(x-y-z)^2=(2x+2y+2z+3x-3y-3z)(2x+2y+2z-3x+3y+3z)
7.(a-3)^2-(2a-6)=(a-3)^2-2(a-3)=(a-3-2)(a-3)=(a-5)(a-3)
2.(x+z)^2-(2x-3)^2=(x+z+2x-3)(x+z-2x+3)=(3x+z-3)(z-x+3)
3.令x-y=a,则原式为9a^2-6a+1=(3a-1)^2=(3x-3y-1)^2
4.16x^4y^4-1=(4^2y^2+1)(4^2y^2-1)=(4^2y^2+1)(2xy+1)(2xy-1)
5.-(m-n)^2+1=1-(m-n)^2=(1+m-n)(1-m+n)
6.4(x+y+z)^2-9(x-y-z)^2=(2x+2y+2z+3x-3y-3z)(2x+2y+2z-3x+3y+3z)
7.(a-3)^2-(2a-6)=(a-3)^2-2(a-3)=(a-3-2)(a-3)=(a-5)(a-3)
看了因式分解a^2b^3-4a^2...的网友还看了以下:
23.一棵前序序列为1,2,3,4的二叉树,其中序序列不可能是().1.一棵前序序列为1,2,3, 2020-07-08 …
求1+2+2^2+2^3+2^4+…+2^2014的值.设S=1+2+2^2+2^3+2^4+…+ 2020-07-09 …
1.计算:1^2+1^4+6^2+7^2=102,2^2+3^2+5^2+8^2=102所以=2. 2020-07-17 …
根式计算化简1、(1/x^2-3x+2)+(1/x^2-x)+(1/x^2+x)+(1/x^2+3 2020-07-30 …
我们规定一种运算“△”,a△b=ab-1,同时规定运算“△”和乘方属于同级运算.如2×3△4=2× 2020-07-31 …
已知矩阵A=(α1α2α3α4)经初等行变换,化为111301120011,则必有()A.α4=α 2020-08-02 …
空间解析几何问题求与两直线(x+3)/2=(y-5)/3=(z-2)/4和x=(y-2)/2=(z 2020-08-02 …
1.|x+2|+|y-3|+(-z+4)^2=0,则x^y+x^z=(只填空,2.已知x>3,则|1 2020-11-01 …
求一道预备班数学期中考试的答案小明在做题时发现了一个规律:1*2/1=1-2/1,2*3/1=2/1 2020-11-05 …
计算,并写出过程.[1]1+7分之1-[-7分之3][2]2.5-4+[-2分之1][3]-3分之1 2020-11-07 …