- wpBeta
-
1.A2-B2+B-1/4=A2-(B2-B+1/4)=A2-(B-1/2)2=(A+B-1/2)(A-B+1/2)
2.4(X2+Y)(X2-Y)-(2X2-Y)2=4(X4-Y2)-(4X4-4X2*Y+Y2)=-5Y2+4X2*Y
代入数值算一下就好咯
- 阿啵呲嘚
-
1)a^2-b^2+b-1/4
=a^2-(b^2-b+1/4)
=a^2-(b-1/2)^2
=(a+b-1/2)(a-b+1/2)
完全平方公式
2)4(x^2+y)(x^2-y)-(2x^2-y)^2
=4(x^4-y^2)-(4x^4-4x^2y+y^2)
=-4y^2+4x^2y-y^2
=4x^2y-5y^2
=100*(-5/2)-25/4
=-250-25/4
=-1025/4
- 南yi
-
A²-B²+B-1/4
=A²-(B²-B+1/4)
=A²-(B-1/2)²
=(A+B-1/2)(A-B+1/2)
4(X²+Y)(X²-Y)-(2X²-Y)²
=4(X^4-Y²)-(4X^4-4X²Y+Y²)
=4X^4-4Y²-4X^4+4X²Y-Y²
=4X²Y-5Y²
=4*4*(-5)-5*25
=-205
- FinCloud
-
(1)a²-b²+b-1/4==A²-(B²-B+1/4)
=A²-(B-1/2)²
=(A+B-1/2)(A-B+1/2)
(2)=4(x²+y)(x²-y)-(2x²-y)²
= 4x^4-4y²-(2x²-y)²
=(2x²)²-(2x²-y)²-4y²
=(2x²+2x²-y)(2x²-2x²+y)-4y²
=(4x²-y)y-4y²
=y(4x²-5y)
=-5×(16+25)
=-205
- 陶小凡
-
1.解:原式=m(m-n)²-n(n-m)²
=m(m-n)²-n(m-n)²
=
(m-n)(m-n)²
=(m-n)³
2.解:原式=(3m+2n)²-(m-n)²
=(3m+2n+m-n)(3m+2n-m+n)
=(4m-n)(2m+3n)
3.解:原式=x³-15x²y-16xy²
=x(x²-15xy-16y²)
=x(x-16y)(x+y)
4.解:原式=-y³-y²-4分之1y
=y(y²-y-1/4)
=y(y-1/2)²
5.解:原式=3分之1x-3x³
=1/3x(1-9x²)
=1/3x(1+3x)(1-3x)
- 里论外几
-
(1)A
(2)D
(3)解:原式=(x-y)(3a-2b)(3a+2b)
(4)解:原式=(a²+2ab+b²-1)(a²-2ab+b²-1)
=[﹙a+b﹚²-1][﹙a-b﹚²-1]
=
﹙a+b+1﹚﹙a+b-1﹚﹙a-b+1﹚﹙a-b-1﹚
- CarieVinne
-
n(n+1)(n+2)(n+3)+1
=n^4+6n^3+11n^2+6n+1(拆解)
=n^4+3n^2+n^2+3n^3+9n^2+3n+n^2+3n+1(重组)
=n^2(n^2+3n+1)+3n(n^2+3n+1)+(n^2+3n+1)(寻找公因式)
=(n^2+3n+1)(n^2+3n+1)(提取公因式)
=(n^2+3n+1)^2(整理)
- 贝贝
-
n(n+1)(n+2)(n+3)+1
=(n+1)(n+2)[n(n+3)]+1
=(n*n+3n+2)(n*n+3n)+1
=(n*n+3n)*(n*n+3n)+2(n*n+3n)+1
=(n*n+3n+1)平方
- max笔记
-
应该是:x²-2x-3=(x+1)(x-3)
- LocCloud
-
(1)16x²+24x+9
(2)-x²+4xy-4y²
(3)m²-14m+49
(4)(m+n)8-4m(m+n)+4m8
(5)x²-5x+6
(6)6x²-7x+2
(7)12x²-11x-15
(8)-6x²+12-x
(9)x²-3xy-10y²+x+9y-2
(10)4x²-14xxy+6y²-7x+y-2