初三一元二次计算1.x²+4x-9=2x-112.x(x+4)=8x+12按照这样的格式写出过程4x²-6x-3=04x²-6x=34x²-6x+3²=3+6(4x-3)²=94x-3=±3x1=9 x2=-153.已知两个连续奇数得积为899,求这两个数4.(例题
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![初三一元二次计算1.x²+4x-9=2x-112.x(x+4)=8x+12按照这样的格式写出过程4x²-6x-3=04x²-6x=34x²-6x+3²=3+6(4x-3)²=94x-3=±3x1=9 x2=-153.已知两个连续奇数得积为899,求这两个数4.(例题](/uploads/image/z/7861056-24-6.jpg?t=%E5%88%9D%E4%B8%89%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E8%AE%A1%E7%AE%971.x%26sup2%3B%2B4x-9%3D2x-112.x%EF%BC%88x%2B4%EF%BC%89%3D8x%2B12%E6%8C%89%E7%85%A7%E8%BF%99%E6%A0%B7%E7%9A%84%E6%A0%BC%E5%BC%8F%E5%86%99%E5%87%BA%E8%BF%87%E7%A8%8B4x%26sup2%3B-6x-3%3D04x%26sup2%3B-6x%3D34x%26sup2%3B-6x%2B3%26sup2%3B%3D3%2B6%EF%BC%884x-3%EF%BC%89%26sup2%3B%3D94x-3%3D%C2%B13x1%3D9+x2%3D-153.%E5%B7%B2%E7%9F%A5%E4%B8%A4%E4%B8%AA%E8%BF%9E%E7%BB%AD%E5%A5%87%E6%95%B0%E5%BE%97%E7%A7%AF%E4%B8%BA899%2C%E6%B1%82%E8%BF%99%E4%B8%A4%E4%B8%AA%E6%95%B04.%EF%BC%88%E4%BE%8B%E9%A2%98)
初三一元二次计算1.x²+4x-9=2x-112.x(x+4)=8x+12按照这样的格式写出过程4x²-6x-3=04x²-6x=34x²-6x+3²=3+6(4x-3)²=94x-3=±3x1=9 x2=-153.已知两个连续奇数得积为899,求这两个数4.(例题
初三一元二次计算
1.x²+4x-9=2x-11
2.x(x+4)=8x+12
按照这样的格式写出过程
4x²-6x-3=0
4x²-6x=3
4x²-6x+3²=3+6
(4x-3)²=9
4x-3=±3
x1=9 x2=-15
3.已知两个连续奇数得积为899,求这两个数
4.(例题)x²-(m-5)x-3m+6
原式=x²-(m-5)x-3(m-2)
=x²-(m-5)x+3*(-m+2)
=(x+3)(x-m+2)
(2)(例题)x²-3x-a²+a+2
原式=x²-3x-(a²-a-2)
=x²-3x-(a+1)(a-2)
=x²-3x+(-a-1)(x+a-2)
(3)x²-(2a+1)x+a²+a
(4)x²+7x-k²-k+12
初三一元二次计算1.x²+4x-9=2x-112.x(x+4)=8x+12按照这样的格式写出过程4x²-6x-3=04x²-6x=34x²-6x+3²=3+6(4x-3)²=94x-3=±3x1=9 x2=-153.已知两个连续奇数得积为899,求这两个数4.(例题
1.x²+4x-9=2x-11
x²+2x=-2
x²+2x+1=-2+1
(x+1)²=-1
因为(x+1)²>=0
所以方程无解
2.x(x+4)=8x+12
x^2+4x=8x+12
x^2-4x=12
x^2-4x+4=12+4
(x-2)^2=16
x-2=±4
x1=6,x2=-2
x²-(2a+1)x+a²+a
=x²-(2a+1)x+a(a+1)
=(x-a)(x-a-1)
x²+7x-k²-k+12
=x²+7x-(k²+k-12)
=x²+7x-(k+4)(k-3)
=[x+(k+4)][x-(k-3)]
=(x+k+4)(x-k+3)
1.x²+2x+2=o
(x+1)²+1=0这题题目错了,是高中复数运算,不是初三了。
2.x²—4x—12=0
(x-2)²-16=0
x-2=±4
x=6或x=-2我有事咯,呵呵
3.(2n-1)(2n+1)=899
4n²-1=899
得n=±25