已知(a²+b²+1)(a²+b²-3)=5求a²+b²的值
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已知(a²+b²+1)(a²+b²-3)=5求a²+b²的值
已知(a²+b²+1)(a²+b²-3)=5求a²+b²的值
已知(a²+b²+1)(a²+b²-3)=5求a²+b²的值
把a²+b²看成整体
令a²+b²=x
则(x+1)(x-3)=5
x²-2x-3-5=0
x²-2x-8=0
(x-4)(x+2)=0
x1=4
x2=-2
因为a²+b²≥0
所以a²+b²=4