若(a2+b2-3)*(a2+b2+3)=7,则a2+b2=
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若(a2+b2-3)*(a2+b2+3)=7,则a2+b2=
![若(a2+b2-3)*(a2+b2+3)=7,则a2+b2=](/uploads/image/z/20145602-2-2.jpg?t=%E8%8B%A5%28a2%2Bb2-3%29%2A%28a2%2Bb2%2B3%29%3D7%2C%E5%88%99a2%2Bb2%3D)
(a2+b2-3)*(a2+b2+3)=7
(a2+b2)^2-9=7
则(a2+b2)^2=16
则a2+b2=4
(a2+b2)^2-9=7
则(a2+b2)^2=16
则a2+b2=4
若(a2+b2)(a2+b2+1)=12,则a2+b2=
(3a2-3ab+2b2)+(a2+4ab-2b2)=4a2+ab+b2
(a2+b2)(a2+b2-3)-4=0,求a2+b2的值
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若(a2+b2)(a2+b2-4)=12,求a2+b2的值.
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已知a2+b2=5,ab=-3,则代数式4a2-4b2+3ab-2a2+6b2+2ab的值是 .
若2a2-2ab+b2+4a+4=0,则a2+b2=
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