因式分解题十道求解TAT
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/11 10:39:23
因式分解题十道求解TAT
16(a-b)^2-25(a+b)^2
-4a^3b^4+12a^2b^5-16ab^6
2(x+1)+(x+1)(x-1)-(2x-3)(x+1)
(m^2+4m+2)^2-4
16a^4-23a^2b^2+9b^4
5a^2b(x-y)^3-30ab^2(y-x)^2
a^2-2ab+b^2-4
(x^2-2x-2)(x^2-2x+4)+9
4a^2-16
今晚回答出来加分,越早越多,明天要交的,可怜可怜我吧QAQ
16(a-b)^2-25(a+b)^2
-4a^3b^4+12a^2b^5-16ab^6
2(x+1)+(x+1)(x-1)-(2x-3)(x+1)
(m^2+4m+2)^2-4
16a^4-23a^2b^2+9b^4
5a^2b(x-y)^3-30ab^2(y-x)^2
a^2-2ab+b^2-4
(x^2-2x-2)(x^2-2x+4)+9
4a^2-16
今晚回答出来加分,越早越多,明天要交的,可怜可怜我吧QAQ
(1)16(a-b)^2-25(a+b)^2
={4(a-b)+5(a+b)}{4(a-b)-5(a+b)}
=(9a+b)(-a-9b)
=-(9a+b)(a+9b)
(2)-4a^3b^4+12a^2b^5-16ab^6
=-4ab^4*(a^2-3ab+4b^2)
(3)2(x+1)+(x+1)(x-1)-(2x-3)(x+1)
=(x+1){2+(x-1)-(2x-3)}
=(x+1)(4-x)
(4)(m^2+4m+2)^2-4
=(m^2+4m+2-2)(m^2+4m+2+2)
=m(m+4)(m^2+4m+4)
=m(m+1)(m+2)^2
(5)16a^4-23a^2b^2+9b^4
=(4a^2-3b^2)^2 +24a^2b^2-23a^2b^2
=(4a^2-3b^2)^2-a^2b^2
=(4a^2-ab-3b^2)(4a^2+ab-3b^2)
=(4a+3b)(a-b)(4a-3b)(a+b)
(6)5a^2b(x-y)^3-30ab^2(y-x)^2
=5b*a^2*(y-x)^2{(x-y)-6}
(7)a^2-2ab+b^2-4
=(a-b)^2 -2^2
=(a-b+2)(a-b-2)
(8)(x^2-2x-2)(x^2-2x+4)+9
=((x^2-2x+1)-3)((x^2-2x+1)+3)+9
=(x^2-2x+1)^2-3^2+9
=(x^2-2x+1)^2
(9)4a^2-16
=(2a-4)(2a+4)
=4(a-2)(a+2)
={4(a-b)+5(a+b)}{4(a-b)-5(a+b)}
=(9a+b)(-a-9b)
=-(9a+b)(a+9b)
(2)-4a^3b^4+12a^2b^5-16ab^6
=-4ab^4*(a^2-3ab+4b^2)
(3)2(x+1)+(x+1)(x-1)-(2x-3)(x+1)
=(x+1){2+(x-1)-(2x-3)}
=(x+1)(4-x)
(4)(m^2+4m+2)^2-4
=(m^2+4m+2-2)(m^2+4m+2+2)
=m(m+4)(m^2+4m+4)
=m(m+1)(m+2)^2
(5)16a^4-23a^2b^2+9b^4
=(4a^2-3b^2)^2 +24a^2b^2-23a^2b^2
=(4a^2-3b^2)^2-a^2b^2
=(4a^2-ab-3b^2)(4a^2+ab-3b^2)
=(4a+3b)(a-b)(4a-3b)(a+b)
(6)5a^2b(x-y)^3-30ab^2(y-x)^2
=5b*a^2*(y-x)^2{(x-y)-6}
(7)a^2-2ab+b^2-4
=(a-b)^2 -2^2
=(a-b+2)(a-b-2)
(8)(x^2-2x-2)(x^2-2x+4)+9
=((x^2-2x+1)-3)((x^2-2x+1)+3)+9
=(x^2-2x+1)^2-3^2+9
=(x^2-2x+1)^2
(9)4a^2-16
=(2a-4)(2a+4)
=4(a-2)(a+2)