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x 2 3x-40 = 0 delta = 3 2-4×1×(-40)= 169 x =(-3√169)/2 =(-3 13)/2x = 5或-8
方法如下,请参考:
X3x-40 = 0直接代入公式:
x 2 3x-40 = 0 delta = 3 2-4×1×(-40)= 169 x =(-3√169)/2 =(-3 13)/2x = 5或-8
用根式AXBX C = 0,根X = (-B√√ (B-4ac))/(2a)这里把a = 4,b =-6,C = 0代入上式得到X = (6 (-6))/8,即X1 = (6-6)/8 = 0x2 =(。
方法如下,请参考:
X3x-40 = 0直接代入公式:
x 2 3x-40 = 0 delta = 3 2-4×1×(-40)= 169 x =(-3√169)/2 =(-3 13)/2x = 5或-8
用根式AXBX C = 0,根X = (-B√√ (B-4ac))/(2a)这里把a = 4,b =-6,C = 0代入上式得到X = (6 (-6))/8,即X1 = (6-6)/8 = 0x2 =(。
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x/y = ln(xy),
法1, 两边对 x 求导, 注意 y 是 x 的函数,得
(y-xy')/y^2 = (y+xy')/(xy), x(y-xy') = y(y+xy'),
得 dy/dx = y' = (xy-y^2)/(x^2+xy) = (y/x)(x-y)/(x+y)
法2, x/y = ln(xy), 即 x = yln(xy), 记 F = yln(xy) - x
F'x = y[y/(xy)] -1 = y/x - 1, F'y = ln(xy) + y[x/(xy)] = x/y + 1
dy/dx = - F'x/F'y = (1-y/x)/(x/y+1) = (y/x)(x-y)/(x+y)
法1, 两边对 x 求导, 注意 y 是 x 的函数,得
(y-xy')/y^2 = (y+xy')/(xy), x(y-xy') = y(y+xy'),
得 dy/dx = y' = (xy-y^2)/(x^2+xy) = (y/x)(x-y)/(x+y)
法2, x/y = ln(xy), 即 x = yln(xy), 记 F = yln(xy) - x
F'x = y[y/(xy)] -1 = y/x - 1, F'y = ln(xy) + y[x/(xy)] = x/y + 1
dy/dx = - F'x/F'y = (1-y/x)/(x/y+1) = (y/x)(x-y)/(x+y)
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来货时,查看是否有被拆开,防止被混进空瓶子。已有发现过
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