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  • 一,解下列關(guān)于x的方程

    一,解下列關(guān)于x的方程
    1.ab^2x^2-(a^4+b^5)x+a^3b^3=0(ab不等于)
    2.a^2(x^2-1)=a(x^2-3)+3
    3.(2x^2-3x-2)n^2+(1-x^2)m^2=mn(1+x^2)
    4.6x^4-13x^3+12x^2-13x+6=0
    5.2x^4+7x^3-x^2-7x+2=0
    6.(12x-1)(6x-1)(4x-1)(3x-1)=5
    全寫出來的再追加50分
    最好能在明天之前寫好
    第一題為(ab不等于0)
    數(shù)學(xué)人氣:844 ℃時(shí)間:2020-06-07 08:01:06
    優(yōu)質(zhì)解答
    說明:
    (1)懷疑第5題題目應(yīng)該是:2x^4+7x^3-4x^2-7x+2=0,原題目想了很久還分解不出來
    (2)2、3題應(yīng)該分情況討論a、m、n的值,比較麻煩
    (3)gbguo,你的最后那道題好像解得有點(diǎn)問題啊
    1、
    ab^2x^2 - (a^4+b^5)x + a^3b^3 = 0
    (ax - b^3)(b^2x - a^3) = 0
    解得:
    x = b^3 / a,或 x = a^3 / b^2
    2、
    a^2(x^2-1)=a(x^2-3)+3
    (a^2 - a)x^2 = a^2 - 3a + 3
    (1) 若 a = 0 或 a = 1,則無解;
    (2) 若 a > 1 或 a < 0,則:
    a(a-1) > 0,a^2 - 3a + 3 = (a - 1.5)^2 + 3/4 > 0
    所以:
    x^2 = (a^2 - 3a + 3) / [a(a-1)] > 0
    x = ±√{(a^2 - 3a + 3) / [a(a-1)] }
    (3)若 0 < a < 1,則:
    a(a-1) < 0,a^2 - 3a + 3 = (a - 1.5)^2 + 3/4 > 0
    所以:
    x^2 = (a^2 - 3a + 3) / [a(a-1)] < 0
    故x無實(shí)數(shù)解(如果能用復(fù)數(shù)的話,x = ±i√{-(a^2 - 3a + 3) / [a(a-1)] })
    3、
    (2x^2-3x-2)n^2+(1-x^2)m^2=mn(1+x^2)
    ( 2n^2 - m^2 - mn )x^2 - 3n^2 * x + (m^2 - 2n^2 - mn) = 0
    (2n + m)(n - m)x^2 - 3n^2 * x + (m - 2n)(m + n) = 0
    ( (2n+m)x + (m+n) )((n-m)x + (m-2n) ) = 0
    (1)若n = m = 0,則:x為任意值
    (2)若n、m不同時(shí)為0,則:
    ① 2n = -m 時(shí),x = (2n-m) / (n-m)
    ② n = m 時(shí),x = (-n-m) / (2n+m)
    ③ 2n ≠ -m,且 n ≠ m時(shí),x1 = (2n-m) / (n-m) ,x2 = (-n-m) / (2n+m)
    4、
    6x^4-13x^3+12x^2-13x+6=0
    ( 6x^4 - 13x^3 + 6x^2 ) + (6x^2 - 13x + 6) = 0
    x^2(6x^2 - 13x + 6) + (6x^2 - 13x + 6) = 0
    (x^2 + 1) (6x^2 - 13x + 6) = 0
    因?yàn)椋簒^2 + 1 > 0,所以:
    6x^2 - 13x + 6 = 0
    (2x - 3) (3x - 2) = 0
    x = 2/3 或 x = 3/2
    5、懷疑題目是:
    2x^4+7x^3-4x^2-7x+2=0
    ( 2x^4 + 7x^3 - 2x^2) - ( 2x^2 + 7x - 2 ) = 0
    x^2 ( 2x^2 + 7x - 2 ) - ( 2x^2 + 7x - 2 )= 0
    ( x^2 - 1 )( 2x^2 + 7x - 2 ) = 0
    (x + 1)(x - 1)( 2x^2 + 7x - 2 ) = 0
    x1 = -1,x2 = 1,x3 = (-7+√65)/4,x4 = (-7+√65)/4
    6、
    (12x-1)(6x-1)(4x-1)(3x-1)=5
    (12x-1)(12x-2)(12x-3)(12x-4)=2*3*4*5
    顯然,左邊是4個(gè)公差為1的等差數(shù)列的積,可令:
    12x - 1 = 5
    12x - 2 = 4
    12x - 3 = 3
    12x - 4 = 2
    得:x = 1/2
    也可以令:
    12x - 1 = -2
    12x - 2 = -3
    12x - 3 = -4
    12x - 4 = -5
    則得:x = -1/12
    為了證明x只有這2個(gè)解,我們令 y = 12x - 3
    則:(y-1)y(y+1)(y+2) = 120 有2個(gè)根y=3、y=-4
    展開,得:
    y^4 + 2y^3 - y^2 - 2y - 120 = 0
    一定可分解為:
    (y + 4)(y - 3)f(y) = 0
    其中 f(y)是關(guān)于 y 的二次函數(shù).
    f(y) = ( y^4 + 2y^3 - y^2 - 2y - 120 )/ (y + 4)(y - 3)
    = ( y^4 + 2y^3 - y^2 - 2y - 120 )/ (y^2 + y - 12)
    = y^2 + (y^3 + 11y^2 - 2y - 120 )/ (y^2 + y - 12)
    = y^2 + y + (10y^2 - 10y - 120) / (y^2 + y - 12)
    = y^2 + y + 10
    = (y + 1/2)^2 + 9.75 > 0
    故 y 只有2根 4和-3
    所以x也只有2根 1/2 和 -1/12
    我來回答
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