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计算题哈,
若代数式x^2+y^2-14x+2y+50的值为0.
化简求值:[2x^2-(x+y)(x-y)]-[(-x-y)(-x+y)+2y^2]
请把每一步写清楚了,
人气:414 ℃ 时间:2020-01-28 12:27:13
解答
如果使代数式x^2+y^2-14x+2y+50的值为0,则x^2+y^2-14x+2y+50=0 (x-7)^2+(y+1)^2=0 x=7,y=-1 [2x^2-(x+y)(x-y)]-[(-x-y)(-x+y)+2y^2] =[2x^2-(x^2-y^2)]-[(x^2-y^2)+2y^2] =(x^2+y^2)-(x^2+y^2)=x^2+y^2-x^2-y^2=0(好...
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