若a-b=5,ab=-6,求:(1)a²+b².(2)a²-b².(3)a
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若a-b=5,ab=-6,求:(1)a²+b².(2)a²-b².(3)a³+b³.
![若a-b=5,ab=-6,求:(1)a²+b².(2)a²-b².(3)a](/uploads/image/z/18213186-66-6.jpg?t=%E8%8B%A5a-b%3D5%2Cab%3D-6%2C%E6%B1%82%3A%281%29a%26%23178%3B%2Bb%26%23178%3B.%282%29a%26%23178%3B-b%26%23178%3B.%283%29a%26%231)
a-b=5,ab=-6
(1)a²+b²
=a²-2ab+b²+2ab
=(a-b)²+2ab
=5²+2×(-6)
=13
(2)(a+b)²
=a²+2ab+b²
=13-2×6
=1
∴a+b=±1
a²-b²
=(a+b)(a-b)
=±1×5
=±5
(3)a³+b³.
=(a+b)(a²-ab+b²)
=±1×﹙13+6)
=±19
(1)a²+b²
=a²-2ab+b²+2ab
=(a-b)²+2ab
=5²+2×(-6)
=13
(2)(a+b)²
=a²+2ab+b²
=13-2×6
=1
∴a+b=±1
a²-b²
=(a+b)(a-b)
=±1×5
=±5
(3)a³+b³.
=(a+b)(a²-ab+b²)
=±1×﹙13+6)
=±19
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