1.(x+3)(2x-m)的积中不含一次项,则常数m=_____,计算结果是_____ 2.计算:(2x^2-1)(x-4)-(x^2+3)(2x-5) 3.若x^3-6x^2+11x-6=(x-1)(x^2+mx+n),求m,n的值 4.解方程:2x(x-3)-(x+6)(x-3)=x^2+12 5.已知a,b,c,d是四个连续的奇数,设
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![1.(x+3)(2x-m)的积中不含一次项,则常数m=_____,计算结果是_____ 2.计算:(2x^2-1)(x-4)-(x^2+3)(2x-5) 3.若x^3-6x^2+11x-6=(x-1)(x^2+mx+n),求m,n的值 4.解方程:2x(x-3)-(x+6)(x-3)=x^2+12 5.已知a,b,c,d是四个连续的奇数,设](/uploads/image/z/6138470-38-0.jpg?t=1.%28x%2B3%29%282x-m%29%E7%9A%84%E7%A7%AF%E4%B8%AD%E4%B8%8D%E5%90%AB%E4%B8%80%E6%AC%A1%E9%A1%B9%2C%E5%88%99%E5%B8%B8%E6%95%B0m%3D_____%2C%E8%AE%A1%E7%AE%97%E7%BB%93%E6%9E%9C%E6%98%AF_____+2.%E8%AE%A1%E7%AE%97%EF%BC%9A%EF%BC%882x%5E2-1%29%28x-4%29-%28x%5E2%2B3%29%282x-5%29+3.%E8%8B%A5x%5E3-6x%5E2%2B11x-6%3D%28x-1%29%28x%5E2%2Bmx%2Bn%29%2C%E6%B1%82m%2Cn%E7%9A%84%E5%80%BC+4.%E8%A7%A3%E6%96%B9%E7%A8%8B%EF%BC%9A2x%28x-3%29-%28x%2B6%29%28x-3%29%3Dx%5E2%2B12+5.%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%2Cd%E6%98%AF%E5%9B%9B%E4%B8%AA%E8%BF%9E%E7%BB%AD%E7%9A%84%E5%A5%87%E6%95%B0%2C%E8%AE%BE)
1.(x+3)(2x-m)的积中不含一次项,则常数m=_____,计算结果是_____ 2.计算:(2x^2-1)(x-4)-(x^2+3)(2x-5) 3.若x^3-6x^2+11x-6=(x-1)(x^2+mx+n),求m,n的值 4.解方程:2x(x-3)-(x+6)(x-3)=x^2+12 5.已知a,b,c,d是四个连续的奇数,设
1.(x+3)(2x-m)的积中不含一次项,则常数m=_____,计算结果是_____
2.计算:(2x^2-1)(x-4)-(x^2+3)(2x-5)
3.若x^3-6x^2+11x-6=(x-1)(x^2+mx+n),求m,n的值
4.解方程:2x(x-3)-(x+6)(x-3)=x^2+12
5.已知a,b,c,d是四个连续的奇数,设其中最小的奇数为d=2n-1(n为正整数),当ac-bd=88时,求出这四个奇数
6.已知把(x^2-x+1)^6展开后得a12x^12+a11x^11+ …… +a2x^2+a1x+a0,求a12+a10+a8+a6+a4+a2+a0的值 (a后面的数字是特指第几个a,本来那些数字在作业本上是很小个的,跟幂一样大小)
能做多少就多少啊,
1.(x+3)(2x-m)的积中不含一次项,则常数m=_____,计算结果是_____ 2.计算:(2x^2-1)(x-4)-(x^2+3)(2x-5) 3.若x^3-6x^2+11x-6=(x-1)(x^2+mx+n),求m,n的值 4.解方程:2x(x-3)-(x+6)(x-3)=x^2+12 5.已知a,b,c,d是四个连续的奇数,设
1 m=6 2x^2-18
2 -3x^2-7x+19
3 m=-5 n=6
4 x=2/3
5 19 21 23 25
1.m=6x=-3 jieguo=0
1. M=6 结果x^2-9