Question

两道【初一】因式分解题目

1.x²y²-x²-y²+4xy+1 因式分解
2.是一个分数方程
(X-3分之X-2)=4-(X²-6X+9分之2)
把过程写下来，会用Word打分数的尽量打出来，过程麻烦各位详细一点，谢谢，我会追加的。

Answer 1

1.
x^2y^2-x^2-y^2+4xy+1
=(x^2y^2-2xy+1)-(x^2-2xy+y^2)
=(xy-1)^2-(x-y)^2
=(xy-x+y)(xy+x-y)

2.
(x-2)/(x-3)=4-2/(x^2-6x+9)
(x-2)/(x-3)=4-2/(x-3)^2
x≠3
同时乘(x-3)^2,得：
(x-2)(x-3)=4(x-3)^2-2
x^2-5x+6=4x^2-24x+36-2
3x^2-19x+28=0
(x-4)(3x-7)=0
x=4或x=7/3

Answer 2

1、
原式=(x²y²+2xy+1)-(x²-2xy+y²)
=(xy+1)²-(x-y)²
=(xy+x-y+1)(xy-x+y+1)

2、
(x-2)/(x-3)=4-2/(x-3)²
两边乘(x-3)²
(x-2)(x-3)=4(x-3)²-2
x²-5x+6=4x²-24x+36-2
3x²-19x+28=0
(3x-7)(x-4)=0
x=7/3,x=4
分式方程要检验
经检验，x=7/3和x=4是方程的解

Answer 3

x²y²-x²-y²+4xy+1
=(x²y²+2xy+1)-(x²+y²-2xy)
=(xy+1)²-(x-y)²
=(xy+1+x-y)(xy+1-x+y)

(x-2)/(x-3)=4-2/(x²-6x+9)
(x-2)(x-3)=4(x²-6x+9)-2
3x²-19x+28=0
(3x-7)(x-4)=0
x1=7/3,x2=4