4(2x^2-x+1)(x^2-2x+3)-(3x^2-3x+4)^2
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4(2x^2-x+1)(x^2-2x+3)-(3x^2-3x+4)^2
我下午就要用
我下午就要用
![4(2x^2-x+1)(x^2-2x+3)-(3x^2-3x+4)^2](/uploads/image/z/15342755-59-5.jpg?t=4%282x%5E2-x%2B1%29%28x%5E2-2x%2B3%29-%283x%5E2-3x%2B4%29%5E2)
4(2x^2-x+1)(x^2-2x+3)-(3x^2-3x+4)^2 =4(2x^2-x+1)(x^2-2x+3)-[(2x^2-x+1)+(x^2-2x+3)]^2 =4(2x^2-x+1)(x^2-2x+3)-(2x^2-x+1)^2-(x^2-2x+3)^2-2(2x^2-x+1)(x^2-2x+3) =-(2x^2-x+1)^2-(x^2-2x+3)^2+2(2x^2-x+1)(x^2-2x+3) =-[(2x^2-x+1)^2+(x^2-2x+3)^2-2(2x^2-x+1)(x^2-2x+3) =-[(2x^2-x+1)-(x^2-2x+3)]^2 =-(x^2+x-2)^2 =-(x+2)^2(x-1)^2
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