1
kx²+(k+1)(k-1)x+k²-k-1=0
kx²+k²x-x+k²-k-1=0
(k²-1)(x+1)+kx²-k=0
(k²-1)(x+1)+k(x+1)(x-1)=0
(x+1)[(k²-1)+k(x-1)]=0
使x+1=0
x= -1
2
4ab-a²+4(1+b)(1-b)=0
4ab-a²+4-4b²=0
a²-4ab+4b²=4
(a-2b)²=4
a-2b=2①
a-2b= -2②
由题目可知
2a+2b=10 ③
联合①和③
a=4,b=1
联合②和③
a=8/3,b=7/3(舍去)
所以
S长方形=ab=4