sat2数学几道高次方程难题,说接这类题方法就行

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  • = C₈⁴ = 8*7*6*5/(1*2*3*4) = 70

    f'(x) = 18x² - 10x + 4,恒为正,即f(x)为单调递增

    f(0) = -15 < 0

    f(1) = -10 < 0

    f(2) = 21 > 0

    有一个根在1,2之间,容易试出是x = 3/2

    f(x) = (2x + 3)(3x² + 2x + 5)

    显然后者无实数解,即f(x) = 0只有一个实数解

    f'(x) = -21x^6 + 10x⁴ - 3

    f"(x) = -126x^5 + 40x^3 = 2x^3(20 - 63x²

    易证x² = 20/63时,f'(x)取最大值,且此最大值小于0,f(x)单调递减增