第一题:x²-x+2-2(x²-x-1)=( )第二题:若使(ax²-2xy+y²)-(-x²+bxy+2y²)=5x²-9xy+cy²永远成立,求a,b,c
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![第一题:x²-x+2-2(x²-x-1)=( )第二题:若使(ax²-2xy+y²)-(-x²+bxy+2y²)=5x²-9xy+cy²永远成立,求a,b,c](/uploads/image/z/7207599-39-9.jpg?t=%E7%AC%AC%E4%B8%80%E9%A2%98%EF%BC%9Ax%26%23178%3B-x%2B2-2%28x%26%23178%3B-x-1%29%3D%28+%29%E7%AC%AC%E4%BA%8C%E9%A2%98%EF%BC%9A%E8%8B%A5%E4%BD%BF%28ax%26%23178%3B-2xy%2By%26%23178%3B%29-%28-x%26%23178%3B%2Bbxy%2B2y%26%23178%3B%29%3D5x%26%23178%3B-9xy%2Bcy%26%23178%3B%E6%B0%B8%E8%BF%9C%E6%88%90%E7%AB%8B%2C%E6%B1%82a%2Cb%2Cc)
第一题:x²-x+2-2(x²-x-1)=( )第二题:若使(ax²-2xy+y²)-(-x²+bxy+2y²)=5x²-9xy+cy²永远成立,求a,b,c
第一题:x²-x+2-2(x²-x-1)=( )
第二题:若使(ax²-2xy+y²)-(-x²+bxy+2y²)=5x²-9xy+cy²永远成立,求a,b,c
第一题:x²-x+2-2(x²-x-1)=( )第二题:若使(ax²-2xy+y²)-(-x²+bxy+2y²)=5x²-9xy+cy²永远成立,求a,b,c
第二题合并同类项啊 (ax²-2xy+y²)-(-x²+bxy+2y²)
=x²(a+1)-(2+b)xy-y²
=5x²-9xy+cy²
对应系数相等 a=4,b=7,c=-1
1,x²-x+2-2(x²-x-1)=(-x²+x+4 )
2,(ax²-2xy+y²)-(-x²+bxy+2y²)=ax²-2xy+y²+x²-bxy-2y²
ax²-2xy+y²+x²-bxy-2y²=5x²...
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1,x²-x+2-2(x²-x-1)=(-x²+x+4 )
2,(ax²-2xy+y²)-(-x²+bxy+2y²)=ax²-2xy+y²+x²-bxy-2y²
ax²-2xy+y²+x²-bxy-2y²=5x²-9xy+cy²
ax²-2xy+y²+x²-bxy-2y²-5x²+9xy-cy²=(a-4)x²+(7-b)xy-(c+1)y²=0 永远成立,则
a=4,b=7,c= -1
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(ax²-2xy+y²)-(-x²+bxy+2y²)=5x²-9xy+cy²
(a+1)x²-(2+b)xy-y²=5x²-9xy+cy²
等式恒成立,则对应同类项系数相等
即:a+1=5,2+b=9,-1=c
得:a=4,b=7,c=-1
祝你开...
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(ax²-2xy+y²)-(-x²+bxy+2y²)=5x²-9xy+cy²
(a+1)x²-(2+b)xy-y²=5x²-9xy+cy²
等式恒成立,则对应同类项系数相等
即:a+1=5,2+b=9,-1=c
得:a=4,b=7,c=-1
祝你开心!希望能帮到你,如果不懂,请Hi我,祝学习进步!O(∩_∩)O
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