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A.maths
發問:
x^3 - (/2 + 1)x^2 + (/2 -k)x+k=0 where k is a constant. (a) prove that x =1 is a root of the equation.
最佳解答:
x3 - (/2 + 1)x2 + (/2 -k)x+k=0 where k is a constant. (a) prove that x =1 is a root of the equation 請問 /2 是否等於√2 設 f(x) = x3 - (√2 + 1)x2 + (√2 -k)x+k 則只需將 x = 1 代入 f(x)，若 f(1) = 0，則x = 1 便是x3 - (√2 + 1)x2 + (√2 -k)x+k = 0 的根 f(x) = 13 - (√2 + 1)12 + (√2 -k)1+k = 1 - √2 – 1 + √2 – k + k = 0 因 f(1) = 0，所以 x = 1 是x3 - (√2 + 1)x2 + (√2 -k)x+k = 0 的根。
其他解答:
A.maths
發問:
x^3 - (/2 + 1)x^2 + (/2 -k)x+k=0 where k is a constant. (a) prove that x =1 is a root of the equation.
最佳解答:
x3 - (/2 + 1)x2 + (/2 -k)x+k=0 where k is a constant. (a) prove that x =1 is a root of the equation 請問 /2 是否等於√2 設 f(x) = x3 - (√2 + 1)x2 + (√2 -k)x+k 則只需將 x = 1 代入 f(x)，若 f(1) = 0，則x = 1 便是x3 - (√2 + 1)x2 + (√2 -k)x+k = 0 的根 f(x) = 13 - (√2 + 1)12 + (√2 -k)1+k = 1 - √2 – 1 + √2 – k + k = 0 因 f(1) = 0，所以 x = 1 是x3 - (√2 + 1)x2 + (√2 -k)x+k = 0 的根。
其他解答:
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