Question

(3x^2-4x+1)(3x^2+4x+1)= 2(3-5a)^2-5(3a-7)(3a+7)= 过程详细一下、谢谢

Answer 1

(3x^2-4x+1)(3x^2+4x+1)
=[(3x^2+1)-4x][(3x^2+1)+4x]
=(3x^2+1)^2-16x^2
=9x^4+6x^2+1-16x^2
=9x^4-10x^2+1

2(3-5a)^2-5(3a-7)(3a+7)
=2(5a-3)^2-5(3a-7)(3a+7)
=50a^2-60a+18-5(9a^2-49)
=50a^2-60a+18-45a^2+245
=50a^2-45a^2-60a+18+245
=5a^2-60a+263

追问

我就是这样做、可是老师打了个错。

追答

题目的类型

追问

就是化简~ 已知xy=8满足x^2y-xy^2-x+y=36.求x^2+y^2的值 详细说下。

追答

x^2y-xy^2-x+y=36
xy(x-y)-(x-y)=36
(xy-1)(x-y)=36
(8-1)(x-y)=36
7(x-y)=36
x-y=36/7

(x-y)^2=1296/49
x^2+y^2-2xy=1296/49
x^2+y^2-2*8=1296/49
x^2+y^2-16=1296/49
x^2+y^2=1296/49+16
x^2+y^2=2080/49

追问

（xy-1)(x-y)=36
（xy-1）怎样得的？

追答

xy(x-y)-(x-y)=36提取公因式 (x-y)
(xy-1)(x-y)=36