小弟刚自学,不懂?题如下(x^2-2x)^2-7(x^2-2x) +12x^2-(p^2+q^2)x+pq(p+q)(p-q)x^2-2xy-8y^2-x-14y-6已知a+b+c+d=0 a^3+b^3+c^3+d^3=3求abc+bcd+cda+dad的值以上题给个过程回答的人谢谢阿都是关于分解因式的
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/29 21:52:25
![小弟刚自学,不懂?题如下(x^2-2x)^2-7(x^2-2x) +12x^2-(p^2+q^2)x+pq(p+q)(p-q)x^2-2xy-8y^2-x-14y-6已知a+b+c+d=0 a^3+b^3+c^3+d^3=3求abc+bcd+cda+dad的值以上题给个过程回答的人谢谢阿都是关于分解因式的](/uploads/image/z/5334751-55-1.jpg?t=%E5%B0%8F%E5%BC%9F%E5%88%9A%E8%87%AA%E5%AD%A6%2C%E4%B8%8D%E6%87%82%3F%E9%A2%98%E5%A6%82%E4%B8%8B%EF%BC%88x%5E2-2x%EF%BC%89%5E2-7%28x%5E2-2x%29+%2B12x%5E2-%28p%5E2%2Bq%5E2%29x%2Bpq%28p%2Bq%29%28p-q%29x%5E2-2xy-8y%5E2-x-14y-6%E5%B7%B2%E7%9F%A5a%2Bb%2Bc%2Bd%3D0+a%5E3%2Bb%5E3%2Bc%5E3%2Bd%5E3%3D3%E6%B1%82abc%2Bbcd%2Bcda%2Bdad%E7%9A%84%E5%80%BC%E4%BB%A5%E4%B8%8A%E9%A2%98%E7%BB%99%E4%B8%AA%E8%BF%87%E7%A8%8B%E5%9B%9E%E7%AD%94%E7%9A%84%E4%BA%BA%E8%B0%A2%E8%B0%A2%E9%98%BF%E9%83%BD%E6%98%AF%E5%85%B3%E4%BA%8E%E5%88%86%E8%A7%A3%E5%9B%A0%E5%BC%8F%E7%9A%84)
小弟刚自学,不懂?题如下(x^2-2x)^2-7(x^2-2x) +12x^2-(p^2+q^2)x+pq(p+q)(p-q)x^2-2xy-8y^2-x-14y-6已知a+b+c+d=0 a^3+b^3+c^3+d^3=3求abc+bcd+cda+dad的值以上题给个过程回答的人谢谢阿都是关于分解因式的
小弟刚自学,不懂?
题如下
(x^2-2x)^2-7(x^2-2x) +12
x^2-(p^2+q^2)x+pq(p+q)(p-q)
x^2-2xy-8y^2-x-14y-6
已知a+b+c+d=0 a^3+b^3+c^3+d^3=3
求abc+bcd+cda+dad的值
以上题给个过程
回答的人谢谢阿
都是关于分解因式的
小弟刚自学,不懂?题如下(x^2-2x)^2-7(x^2-2x) +12x^2-(p^2+q^2)x+pq(p+q)(p-q)x^2-2xy-8y^2-x-14y-6已知a+b+c+d=0 a^3+b^3+c^3+d^3=3求abc+bcd+cda+dad的值以上题给个过程回答的人谢谢阿都是关于分解因式的
(x^2-2x)^2-7(x^2-2x)+12
=(x^2-2x-3)(x^2-2x-4)
=(x-3)(x+1)(x^2-2x-4)
2.
pq(p^2-q^2)=pq(p-q)(p+q)=(p^2-pq)(pq+q^2)
x^2-(p^2+q^2)x+(p^2-pq)(pq+q^2)
=[x-(p^2-pq)][x-(pq+q^2)
3.
解法一
用双十字相乘法分解
x^2-2xy-8y^2-x-14y-6
=(x^2-2xy-8y^2)-x-14y-6
=(x-4y)(x+2y)-x-14y-6
=(x-4y-3)(x+2y+2)
解法二
用待定系数法分解
x^2-2xy-8y^2-14y-6-x
=(x+ay+b)(x+cy+d)
=x^2+(a+c)xy+acy^2+(a+c)y+(b+d)x+bd
a+c=-2
ac=-8
bc+ad=-14
b+d=-1
bd=-6
a=2,b=2,c=-4,d=-1
x^2-2xy-8y^2-x-14y-6
=(x+2y+2)(x-4y-3)
4.
a+b+c+d=0
=>a+b=-(c+d)--------两边3次方
=>(a+b)^3=-(c+d)^3
=>(a+b)^3+(c+d)^3=0
=>a^3+b^3+c^3+d^3+3(a^2b+b^2a+c^2d+d^2c)=0
=>a^3+b^3+c^3+d^3+3[ab(a+b)+cd(c+d)]=0----a+b=-(c+d);
=>a^3+b^3+c^3+d^3=3[ab(c+d)+cd(a+b)]----展开就是结论
=>a^3+b^3+c^3+d^3=3(abc+bcd+cda+abd)
即:abc+bcd+cda+bad=1