(1)以知:a+b=1 ab=-0.5 就a(a+b)*(a-b)-a*(a+b)的2次方 (2) 用简便方程计算.(x-2)*(x-3)+(x-4)*(x-5)+(x-2)*(x-4)+(x-3)*(x-5)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 03:30:09
![(1)以知:a+b=1 ab=-0.5 就a(a+b)*(a-b)-a*(a+b)的2次方 (2) 用简便方程计算.(x-2)*(x-3)+(x-4)*(x-5)+(x-2)*(x-4)+(x-3)*(x-5)](/uploads/image/z/14750167-31-7.jpg?t=%281%29%E4%BB%A5%E7%9F%A5%3Aa%2Bb%3D1+ab%3D-0.5+%E5%B0%B1a%28a%2Bb%29%2A%28a-b%29-a%2A%28a%2Bb%29%E7%9A%842%E6%AC%A1%E6%96%B9+%282%29+%E7%94%A8%E7%AE%80%E4%BE%BF%E6%96%B9%E7%A8%8B%E8%AE%A1%E7%AE%97.%28x-2%29%2A%28x-3%29%2B%28x-4%29%2A%28x-5%29%2B%28x-2%29%2A%28x-4%29%2B%28x-3%29%2A%28x-5%29)
(1)以知:a+b=1 ab=-0.5 就a(a+b)*(a-b)-a*(a+b)的2次方 (2) 用简便方程计算.(x-2)*(x-3)+(x-4)*(x-5)+(x-2)*(x-4)+(x-3)*(x-5)
(1)以知:a+b=1 ab=-0.5
就a(a+b)*(a-b)-a*(a+b)的2次方
(2) 用简便方程计算.
(x-2)*(x-3)+(x-4)*(x-5)+(x-2)*(x-4)+(x-3)*(x-5)
(1)以知:a+b=1 ab=-0.5 就a(a+b)*(a-b)-a*(a+b)的2次方 (2) 用简便方程计算.(x-2)*(x-3)+(x-4)*(x-5)+(x-2)*(x-4)+(x-3)*(x-5)
一
a(a+b)(a-b)-a(a+b)(a+b)
=a(a+b)[(a-b)-(a+b)]
=a(a+b)[a-b-a+-b]
=-2ab(a+b)
=-2*0.5*1
=-1
二
(x-2)(x-3)+(x-4)(x-5)+(x-2)(x-4)+(x-3)(x-5)
=[(x-2)(x-3)+)+(x-2)(x-4)]+[(x-4)(x-5)+(x-3)(x-5)]
={(x-2)[(x-3)+(x-4)]}+{(x-5)[(x-4)+(x-3)]}
=(x-2)(2x-7)+(x-5)(2x-7)
=(2x-7)[(x-2)+(x-5)]
=(2x-7)的平方
分也不多,随便帮帮你啦,不知道来得及吗?
楼上的第一题算错了,哪有你那样解2次方程的呀.
∵a+b=1
∴a=1-b
∵ab=-0.5
∴a=-0.5 1/b
∴ -0.5 1/b=1-b
b=0.5
∴a=1-0.5
=0.5
∴a(a+b)(a-b)-(a+b)(a+b)
= (a+b)[a(a+b)-(a+b)]
∵a=0.5 b=0.5
=(0.5+0.5)[0.5*(...
全部展开
∵a+b=1
∴a=1-b
∵ab=-0.5
∴a=-0.5 1/b
∴ -0.5 1/b=1-b
b=0.5
∴a=1-0.5
=0.5
∴a(a+b)(a-b)-(a+b)(a+b)
= (a+b)[a(a+b)-(a+b)]
∵a=0.5 b=0.5
=(0.5+0.5)[0.5*(0.5+0.5)-(0.5+0.5)
=-0.5
∴a(a+b)(a-b)-(a+b)(a+b)的值为-0.5
(x-2)*(x-3)+(x-4)*(x-5)+(x-2)*(x-4)+(x-3)*(x-5)=
(x-2)*(2x-7)+(x-5)*(2x-7)=(2x-7)^2
收起
(1)原式提公因式a(a+b)得到a(a+b)(a-b-a-b)=a(a+b)(-2b)=-2ab(a+b)=-2*(-0.5)*1=1
(2)原式第一项和第三项,第二项和第四项提取公因式得到
(x-2)*(x-3+x-4)+(x-5)*(x-4+x-3)=(x-2)(2x-7)+(x-5)*(2x-7)=(2x-7)*(x-2+x-5)=(2x-7)*(2x-7)=4x^2-28x+49
1.因式分解=
a(a+b)[(a-b)-(a+b)]=a(a+b)[a-b-a-b]=-2ab(a+b)=-2*(-0.5)*1=1
2.因式分解=(1,3项结合,2,4项结合)
(x-2)(x-3+x-4)+(x-5)(x-4+x-3)=(2x-7)(x-2+x-5)=(2x-7)的平方