整体代入分解因式(x 2y)的平方-2(x 2y) 1
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原式=y(x2-x+14)=y(x-12)2.故答案为:y(x-12)2
解题思路:分解因式解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.ph
1个问题:将y=(1-2x)/3代入4x+5y=22x+3y=14x+5(1-2x)/3=23y=1-2x4x+(5-10x)/3=2y=(1-2x)/312x+5-10x=62x=1x=0.52个问
(x+2y-3xy)-(-2x-y+xy),其中x+y=二分之一,xy=-二分之一用整体代入法求值.
(1)=m(m+4)-1+5=m²+4m+4=(m-2)²(2)=a(a²-b²)=a(a+b)(a-b)
(1)9a-ab2=a(9-b2)=a(3+b)(3-b); (2)3x3-6x2y+3xy2=3x(x2-2xy+y2)=3x(x-y)2;(3)a2(2a-3)+b2(3-2a)=(2a
3x2(a+3)-4x2y(a+3)=x2(a+3)(3-4y),当a=-0.5,x=3,y=1时,原式=32×(-0.5+3)×(3-4×1)=-22.5.
(1)2x2y-8y=2y(x2-4)=2y(x+2)(x-2).(2)(x-y)2-2x+2y+1=(x-y)2-2(x-y)+1=(x-y-1)2.
∵代数式x3+y3+3x2y+axy2含有因式x-y,∴当x=y时,x3+y3+3x2y+axy2=0,∴令x=y,即x3+x3+3x3+ax3=0,则有5+a=0,解得a=-5.将a=-5代入x3+
解题思路:因式分解进行证明解题过程:题中应该是:(a-c)²-4(a-b)(b-c)=0证明:∵(a-c)²-4(a-b)(b-c)=0a²-2ac+c²-4(
a²+2ab+b²=(a+b)²a²-2ab+b²=(a-b)²a²-b²=(a+b)(a-b)
x3-4x2y+4xy2,=x(x2-4xy+4y2),=x(x-2y)2.
原式=y(x2+2x+1)=y(x+1)2,故答案为:y(x+1)2.
解题思路:此题考查了分式的加减法,分式加减法的关键是通分,通分的关键是找最简公分母解题过程:
8x2y-8xy+2y,=2y(4x2-4x+1),=2y(2x-1)2.
解题思路:提取公因式解答解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq
解题思路:利用立方差公式灵活进行分析。解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/includ
①x2y2-5x2y-6x2=x2(y2-5y-6)=x2(y-6)(y+1);②(p2+q2)2-4p2q2=(p2+q2+2pq)(p2+q2-2pq)=(p+q)2(p-q)2;③(a-b)4-
x2y-2xy-y=y(x2-2x-1)=y(x2-2x+1-2)=y[(x-1)2-(2)2]=y(x-1+2)(x-1-2),故答案为:y(x-1+2)(x-1-2).
(1)x2-9y2=(x+3y)(x-3y);(2)2x2y-8xy+8y=2y(x2-4x+4)=2y(x-2)2;(3)m3(a-2)+m(2-a)=m(2-a)(m2+1).