若a1(x-1)∧4+a2(x-1)³+a3(x-1)²+a4(x-1)+a5=x∧4,则a2-a3+a4=多少
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![若a1(x-1)∧4+a2(x-1)³+a3(x-1)²+a4(x-1)+a5=x∧4,则a2-a3+a4=多少](/uploads/image/z/1664103-39-3.jpg?t=%E8%8B%A5a1%EF%BC%88x-1%EF%BC%89%E2%88%A74%2Ba2%EF%BC%88x-1%EF%BC%89%26%23179%3B%2Ba3%EF%BC%88x-1%EF%BC%89%26%23178%3B%2Ba4%EF%BC%88x-1%EF%BC%89%2Ba5%EF%BC%9Dx%E2%88%A74%2C%E5%88%99a2-a3%2Ba4%EF%BC%9D%E5%A4%9A%E5%B0%91)
若a1(x-1)∧4+a2(x-1)³+a3(x-1)²+a4(x-1)+a5=x∧4,则a2-a3+a4=多少
若a1(x-1)∧4+a2(x-1)³+a3(x-1)²+a4(x-1)+a5=x∧4,则a2-a3+a4=多少
若a1(x-1)∧4+a2(x-1)³+a3(x-1)²+a4(x-1)+a5=x∧4,则a2-a3+a4=多少
令x=1,则a5=1
令x=0,则a1-a2+a3-a4+a5=0
再观察等式左侧x的最高次项为a1 x^4,右侧为x^4,则a1=1
因此a2-a3+a4=a1+a5=2