標題:
問代數, thzzz=]
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發問:
想問四條代數(想要埋個過程)=) 1) 6(p+q)^2–2(p+q) 2) 2(x-y)^2–x(x-y) 3) 3(y-x)^2+2(x-y) 4) 2x(x+y)^2–(x+y)^3
最佳解答:
1) 6(p+q)^2–2(p+q) = 2 (p + q) [3 (p + q) - 1] = 2 (p + q) (3p + 3q - 1) 2) 2(x-y)^2–x(x-y) = (x - y) [2 (x - y) - x] = (x - y) (2x - 2y - x) = (x - y) (x - 2y) 3) 3(y-x)^2+2(x-y) = 3 (x - y)^2 + 2 (x - y) = (x - y) [3 (x - y) + 2] = (x - y) (3x - 3y + 2) 4) 2x(x+y)^2–(x+y)^3 = [(x+y) ^ 2] [2x - (x + y)] = [(x+y) ^ 2] (2x - x - y) = (x - y) [(x+y) ^ 2]
其他解答:
1) =6(p+q)(p+q)-2(p+q) =2(p+q)(3p+3q-1) =2(3p^2+3pq-p+3pq+3q^2-q) =2(3p^2+6pq+3q^2-q) =6p^2+12pq+6q^2-2q 2) =2(x-y)(x-y)-x(x-y) =(x-y)(2x-2y-x) =(x-y)(x-2y) =x^2-2xy-xy+2y^2 =x^2-3xy+2y^2 3) =3(x-y)(x-y)+2(x-y) =(x-y)(3x-3y+2x+2y) =(x-y)(5x-y) =5x^2-xy-5xy+y^2 =5x^2-6xy+y^2 4) =2x(x+y)(x+y)-(x+y)(x+y)(x+y) =(x+y)(x+y)(2x-x-y) =(x+y)(x+y)(x-y) =(x^2-y^2)(x+y) =(x^3+x^2y-xy^2-y^3
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