f(x)=x^90-400x^98+400x^97-400x^96+…400x^2+400x-1.Tính f(x)399 23/08/2021 Bởi Julia f(x)=x^90-400x^98+400x^97-400x^96+…400x^2+400x-1.Tính f(x)399
Mình sửa lại lũy thừa đầu tiên là $x^{99}$ $ f(x) = x^{99} – 400x^{98} + 400x^{97} – 400x^{96} + … – 400x^2 + 400x -1$ $\\$ $ = x^{99} – ( 400x^{98} – 400x^{97} + 400x^{96} – … + 400x^2 – 400x +1)$ $\\$ $\to f(399) = x^{99} – [ (x+1).x^{98} – (x+1).x^{97} + (x+1).x^{96} – … + (x+1).x^2 – (x+1).x +1 )$ $\\$ $ \to f(399) = x^{99} – (x^{99} + x^{98} – x^{98} – x^{97} + x^{97} + x^{96} – … + x^3 +x^2 – x^2 -x +1)$ $\\$ $ \to f(399) = x^{99} – (x^{99} – x + 1)$ $\\$ $ \to f(399) = x -1 = 399 -1 = 398$ Bình luận
Với `x = 399 => x – 399 =0` Ta có: `f(x) = x^99 – 400x^98 + 400x^97 – ….- 400x^2 + 400x -1` `f(x) = x^99 – (399+1)x^98 + (399+1)x^97 -…-(399+1)x^2 + (399+1)x -1` `f(x) = x^99 – 399x^98 – x^98 + 399x^97 + x^97 -….-399x^2 – x^2 + 399x + x -1` `f(x) = x^98(x -399) – x^97( x – 399) +….- x(x -399) + x-1` `f(x) = x^98 .0 + x^97.0 + ….+ x.0 + x-1` `f(x) = x-1` `f(399) = 399-1` `f(399) = 398` Vậy `f(399) = 398` Bình luận
Mình sửa lại lũy thừa đầu tiên là $x^{99}$
$ f(x) = x^{99} – 400x^{98} + 400x^{97} – 400x^{96} + … – 400x^2 + 400x -1$
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$ = x^{99} – ( 400x^{98} – 400x^{97} + 400x^{96} – … + 400x^2 – 400x +1)$
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$\to f(399) = x^{99} – [ (x+1).x^{98} – (x+1).x^{97} + (x+1).x^{96} – … + (x+1).x^2 – (x+1).x +1 )$
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$ \to f(399) = x^{99} – (x^{99} + x^{98} – x^{98} – x^{97} + x^{97} + x^{96} – … + x^3 +x^2 – x^2 -x +1)$
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$ \to f(399) = x^{99} – (x^{99} – x + 1)$
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$ \to f(399) = x -1 = 399 -1 = 398$
Với `x = 399 => x – 399 =0`
Ta có: `f(x) = x^99 – 400x^98 + 400x^97 – ….- 400x^2 + 400x -1`
`f(x) = x^99 – (399+1)x^98 + (399+1)x^97 -…-(399+1)x^2 + (399+1)x -1`
`f(x) = x^99 – 399x^98 – x^98 + 399x^97 + x^97 -….-399x^2 – x^2 + 399x + x -1`
`f(x) = x^98(x -399) – x^97( x – 399) +….- x(x -399) + x-1`
`f(x) = x^98 .0 + x^97.0 + ….+ x.0 + x-1`
`f(x) = x-1`
`f(399) = 399-1`
`f(399) = 398`
Vậy `f(399) = 398`