佩如 發問時間: 科學數學 · 10 年前

數學式的運算和聯立方程式附算式~謝謝

1.設a、b為2x-3x+2=0之二根,求(1)a^2+b^2;(2)b/2a+1 + a/2b+1(3)a^3+b^3

2.解方程式(x+2)(x+3)(x+4)(x+5)-120=0之實根?

3.解(x-2)(x+3)(x-4)(x+6)+6x^2=0

4.設二次方程式ax^2+(a-1)x+a=0有二實根,其中a為一實數,則a的範圍為?

5.設x^2-mx+n=0之二根為連續二整數,則m^2-4n之值為?

6.設方程式^2+6x+m=0的一根是另一根的二倍,求m=?

7.設a、b為方程式x^2+x+1=0之二根,則a+b等於?

8.設方程式2x^2+3x+4=0的兩根為a、b,則a^4+b^4的值等於?

9.化簡x^2+2x+1/x-3*x^2+5x-6/x^2-1*x^2-5x+6/x^3+1得?

10.2x+11/(x-2)(x+3)=a/x-2+b/x+3,求a,b值

11.x^3+16/(x-2)^4=A/x-2+B/(x-2)^2+C/(x-2)^3+D/(x-2)^4,求A,B,C,D

12.求x^3-3/(x-2)(x^2+1)之部分分式

13.若a、b均為實數,且a^3=2+√5,b^3=2-√5,求a+b=?

14.解{1/x+1/y+1/z=6

2/x-1/y+3/z=9

1/x+3/y-1/=4

已更新項目:

是這樣才對

14.解{1/x+1/y+1/z=6

2/x-1/y+3/z=9

1/x+3/y-1/=4

3 個解答

評分
  • 10 年前
    最佳解答

    1. (1) a+b=-(-3)/2=3/2,ab= 1

    a^2+b^2=(a+b)^2-2ab=9/4-2=1/4

    (2) b/( 2a +1)+a/(2b+1)

    =[b( 2a +1)+a(2b+1)]/[( 2a +1)(2b+1)]

    =(4ab+a+b)/[4ab+2(a+b)+1]

    =(4+3/2)/(4+3+1)

    =(8+3)/(8+6+2)

    =11/16

    (3) a^3+b^3=(a+b)(a^2+b^2-ab)

    =(3/2)*(1/4-1)

    =(3/2)*(-3/4)

    =-9/8

    2.(x+2)(x+3)(x+4)(x+5)-120=0

    (x+2)(x+5)(x+3)(x+4)-120=0

    (x^2+7x+10)(x^2+7x+12)-120=0

    (x^2+7x)^2+22(x^2+7x)=0

    (x^2+7x)(x^2+7x+22)=0

    x=0 or x=-7為2實根

    3.(x-2) (x+3)(x-4)(x+6)+6x^2=0

    (x-2) (x+6)(x+3)(x-4)+6x^2=0

    (x^2-12+4x)(x^2-2-x)+6x^2=0

    (x^2-12)^2+3x(x^2-12)-4x^2+6x^2=0

    (x^2-12)^2+3x(x^2-12)+2x^2=0

    (x^2-12+x)(x^2-12+2x)=0

    (x-3)(x+4)(x^2+2x-12)=0

    x=3 or x=-4 or x=-1+/-√13

    4.D=(a-1)^2-4*a*a>= 0

    a^2-2a+1-4a^2>=0

    3a ^2+ 2a-1<=0

    ( 3a-1)(a+1)<=0

    (a-1/3)(a+1)<=0

    -1<=a<=1/3

    5.設兩根 a,a+ 1

    a+a+1=m,a(a+1)=n

    m^2-4n=( 2a +1)^2-4a(a+1)= 4a ^2+ 4a +1-4a^2-4a=1

    6.設兩根a,2a

    a+ 2a = 3a =-6

    a=-2

    (x+2)(x+4)=x^2+6x+ 8

    m=8

    7.a+b=-1

    8.a+b=-3/2

    2x^2+3x+4=0

    2x^2=-3x-4

    2a ^2=-3a-4

    2(a^2+b^2)=-3(a+b)-8=-3(-3/2)-8=-7/ 2

    a^4+b^4=(a^2+b^2)^2-2a^b^2=(-7/4)^2-2*(9/4)=-23/16

    9.(x^2+2x+1)/(x-3)*(x^2+5x-6)/(x^2-1)*(x^2-5x+6)/(x^3+1)

    =[(x^2+2x+1)/(x-3)]*[(x^2+5x-6)/(x^2-1)]*[(x^2-5x+6)/(x^3+1)]

    =[(x^2+2x+1)/(x-3)]*[(x-1)(x+6)/(x^2-1)]*[(x-2)(x-3)]/(x^3+1)]

    =(x+1)(x+6)(x-2)/(x^3+1)

    10.(2x+11)/[(x-2)(x+3)]=a /(x-2)+b/(x+3)

    2x+11=a(x+3)+b(x-2)

    2x+11=(a+b)x+( 3a-2b)

    a+b=2,, 3a--2b=11

    ( 3a--2b)+2(a+b)=11+ 4

    a=3,,b=--1

    2011-04-08 10:29:38 補充:

    11.(x^3+16)/(x-2)^4=A/(x-2)+B/(x-2)^2+C/(x-2)^3+D/(x-2)^4

    x^3+16=A(x-2)^3+B(x-2)^2+C(x-2)+D

    when x=2

    8+16=D=24

    x^3-8=A(x-2)^3+B(x-2)^2+C(x-2)

    A=1

    x^3-8=x^3—6x^2+12x-8+B(x-2)^2+C(x-2)

    6x^2-12x=B(x-2)^2+C(x-2)

    6x=B(x-2)+C

    B=6,C=12

    2011-04-08 10:29:57 補充:

    12.(x^3-3)/[(x-2)(x^2+1)]=1+a/(x-2)+(bx+c)/(x^2+1)

    x^3-3=(x-2)(x^2+1)+a(x^2+1)+(bx+c)(x-2)

    x^3-3=x^3-2x^2+x-2+a(x^2+1)+(bx+c)(x-2)

    2x^2-x-1=a(x^2+1)+(bx+c)(x-2)

    when x=2

    8-2-1=a(5)

    a=1

    2x^2-x-1=x^2+1+(bx+c)(x-2)

    x^2-x-2=(bx+c)(x-2)

    x+1=bx+c

    b=1,c=1

    (x^3-3)/[(x-2)(x^2+1)]=1+1/(x-2)+(x+1)/(x^2+1)

    2011-04-08 10:30:21 補充:

    13x=a+b

    a^3b^3=4-5=-1

    ab=-1

    a^3+b^3=4

    (a+b)^3=a^3+b^3+3ab(a+b)=4+3(-1)(a+b)

    x^3=4-3x

    x^3+3x-4=0

    (x^2+x+4)(x-1)=0

    x^2+x+4=0>0

    x=1

    2011-04-08 10:30:58 補充:

    14(1/x+1/y+1/z)+(2/x-1/y+3/z)=6+9

    3/x+4/z=15

    3(1/x+1/y+1/z)-(1/x+3/y-1/z)=3*6-4

    2/x+4/z=14

    (3/x+4/z)-(2/x+4/z)=15-14

    1/x=1

    x=1

    3/1+4/z=15

    z=1/3

    1/y=6-1/x-1/z=6-1-3=2

    y=1/2

  • 10 年前

    1.先生,你的平方去哪了?

    要求ax^2+bx+c=0的兩根之和和兩根之積很簡單

    和=-b/a,積=c/a

    設兩根分別是m和n,則m^2+n^2=(m+n)^2-2mn=(b^2/a^2)-(2c/a)

    m/(2n+1) + n/(2m+1)=[2(m+n)^2-4mn+(m+n)]/[4mn+2(m+n)+1]

    m^3+n^3=(m+n)(m^2-mn+n^2)=(m+n)[(m+n)^2-3mn]

    2.設y=x^2+7x

    則(x+2)(x+5)=y+10

    (x+3)(x+4)=y+12

    乘起來(y+10)(y+12)-120=0

    y^2+22y=0

    y=0or-22

    x^2+7x=0 or 22(22不合)

    X=0 or -7

    其它有空再打

    2011-04-09 08:53:08 補充:

    3.(x-2)(x+3)(x-4)(x+6)+6x^2=0

    (x^2-12+4x)(x^2-12-x)+6x^2=0

    (x^2-12)^2+3x(x^2-12)-4x^2+6x^2=0

    (x^2-12)^2+3x(x^2-12)+2x^2=0

    (x^2+x-12)(x^2+2x-12)=0

    (x+4)(x-3)(x^2+2x-12)=0

    x=-4,3,-1+根號13,-1-根號13

  • YTTSAI
    Lv 5
    10 年前

    請問第14題的第3式 [1/x+3/y-1/=4] 到底1是除以多少?

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