设a²+2a-1=0,b四次方-2b²-1=0,且1-ab²≠0(ab²+b²-3a+1)五次方则求 —————————— 的值 a五次方
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![设a²+2a-1=0,b四次方-2b²-1=0,且1-ab²≠0(ab²+b²-3a+1)五次方则求 —————————— 的值 a五次方](/uploads/image/z/444233-65-3.jpg?t=%E8%AE%BEa%26%23178%3B%2B2a-1%3D0%2Cb%E5%9B%9B%E6%AC%A1%E6%96%B9%EF%BC%8D2b%26%23178%3B-1%3D0%2C%E4%B8%941-ab%26%23178%3B%E2%89%A00%EF%BC%88ab%26%23178%3B%2Bb%26%23178%3B-3a%2B1%EF%BC%89%E4%BA%94%E6%AC%A1%E6%96%B9%E5%88%99%E6%B1%82++%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94%E2%80%94++%E7%9A%84%E5%80%BC++++++++++++++++++++++a%E4%BA%94%E6%AC%A1%E6%96%B9)
设a²+2a-1=0,b四次方-2b²-1=0,且1-ab²≠0(ab²+b²-3a+1)五次方则求 —————————— 的值 a五次方
设a²+2a-1=0,b四次方-2b²-1=0,且1-ab²≠0
(ab²+b²-3a+1)五次方
则求 —————————— 的值
a五次方
设a²+2a-1=0,b四次方-2b²-1=0,且1-ab²≠0(ab²+b²-3a+1)五次方则求 —————————— 的值 a五次方
b^4-2b^-1=0,
两边都除以(-b^4),得
(1/b^)^+2/b^-1=0,
又a^+2a-1=0,ab^≠1,
∴a,1/b^是方程x^+2x-1=0的两根,
∴a/b^=-1,a=-b^,
原式=(b^+b^/a-3+1/a)^5
=(-a-4+1/a)^5
=[(-a^-4a+1)/a]^5
=(-2)^5<...
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b^4-2b^-1=0,
两边都除以(-b^4),得
(1/b^)^+2/b^-1=0,
又a^+2a-1=0,ab^≠1,
∴a,1/b^是方程x^+2x-1=0的两根,
∴a/b^=-1,a=-b^,
原式=(b^+b^/a-3+1/a)^5
=(-a-4+1/a)^5
=[(-a^-4a+1)/a]^5
=(-2)^5
=-32.
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