运用乘法公式:(3x-5)²-(2x 7)²
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1.运用平方差公式,原式=(3x-5+2x+7)(3x-5-2x-7)=(5x+2)(x-12)=5x的平方-58x-242.原式=(x的平方-4)的平方=x的4次方-8x的平方+163.原式=[2x
1、(X-Y-1)(X-Y+5)都有x-y,把它看成一个整体t原式就是(t-1)(t+5)=t^2+4t-5=(x-y)^2+2(x-y)-52、这个问题实际上相邻两个奇数的平方差,奇数可以表示为2n
(x+2y-3z)(x-2y+3z)=[x+(2y-3z)][x-(2y-3z)]=x²-(2y-3z)²=x²-(4y²-12yz+9z²)=x
(3x-2)的平方乘(3x+2)的平方=(9x²-4)²=81x^4-72x²+16
(1)(3x-5)^2-(2x+7)^2=(3x-5+2x+7)(3x-5-2x-7)=(5x+2)(x-12)=5x^2-58x-24(2)a(a-3)-(a-1)=a^2-3a-a+1=a^2-4
1.=[(3X-5)+(2X+7)]*[(3X-5)-(2X+7)]=(7X+2)*(X-12)=7X的平方-82X-242.=[(x+y)+1]*[(x+y)-1]=[(x+y)-1]的平方
(1)(3x-5)²-(2x+7)²=(3x-5+2x+7)(3x-5-2x-7)=(5x+2)(x-12)=5x²-58x-24.(2)(x+y+1)(x+y-1)=(
(3x-5)的平方-(2x+7)的平方=9x²-30x+25-4x²-28x-49=5x²-58x-24=(5x+2)(x-12)
(1)(x-a)²=x²-2ax+a²,此时等于代数式x²-8x+2b可知-2a=-8,a=4;a²=16=2b,b=8因此b-ab=8-32=-24
(3X-5)^2-(2X+7)^2=(3x-5+2x+7)*(3x-5-2x-7)=5x²-58x-24(2X-Y-3)^2=(2x-y)²-2×3(2x-y)+3²=4
(1)(998)2,=(1000-2)2,=1000000-4000+4,=996004;(2)197×203,=(200-3)(200+3),=2002-32,=40000-9,=39991.
(x+2y-3)(x-2y+3)=x²-(2y-3)²=x²-4y²+12y-9(x+5)的平方-(x-2)(x-3)=x²+10x+25-x
1.(3x-5)²-(2x+7)²=9x²-30x+25-(4x²+28x+49)=9x²-30x+25-4x²-28x-49=5x
原式=(x^2-4)^2=x^4-8x^2+16满意请采纳
令2013=a原式=[(a+2)²+a]/[a²-(a+2)]*[(a-1)²-(a-1)]/[a(a+3)-4]=(a²+5a+4)/(a²-a-2
(2x+y-3)²=[2x+(y-3)]²=4x²+4x(y-3)+(y-3)²=4x²+4xy-12x+y²-6y+9
(23a-b)(23a+b)=(23a)2-b2=49a2-b2;(-2x-5)(2x-5)=(-5)2-(2x)2=25-4x2.故答案是:
原式=-(1+3x)²=-9x²-6x-1再问:已知x的3m次方y的2n次方除以(-xy)的5次方等于-xy,则m+n等于多少?再答:3m-5=1,2n-5=1所以m=2,n=3
解题思路:利用平方差公式解题过程:A-B=1234567×1234569-12345682=(1234568-1((1234568+1)-12345682=12345682-1-12345
4.原式=(100-1)×(100+1)×9999=(10000-1)×9999=9999×9999=(10000-1)²=100000000-20000+1=99980001