来自彭巨光的问题
设a2+2a-1=0,b4-2b2-1=0,且1-ab2≠0,则=A.-23B.23C.-32D.32
设a2+2a-1=0,b4-2b2-1=0,且1-ab2≠0,则=
A.-23
B.23
C.-32
D.32
1回答
2020-01-31 21:08
设a2+2a-1=0,b4-2b2-1=0,且1-ab2≠0,则=A.-23B.23C.-32D.32
设a2+2a-1=0,b4-2b2-1=0,且1-ab2≠0,则=
A.-23
B.23
C.-32
D.32
将a2+2a-1=0,b4-2b2-1=0两式相减得:a2-b4+2a+2b2=0,分解因式得:(a+b2)(a-b2+2)=0,若a-b2+2=0,则1-ab2=1-a(a+2)=-(a2+2a-1)=0,而已知1-ab2≠0,所以a-b2+2=0不成立,若a+b2=0,则b2=-a,则=()5=()...