帮忙解几道一元二次方程的难题1.如果方程（x-1)( x² -2x+m)=0的三个实根可以作为一个三角形的三边,求实数m的取值范围.2.是否存在质数p,q,使得关于x的一元二次方程p x² -qx+p=0有有理数根.

帮忙解几道一元二次方程的难题1.如果方程（x-1)( x² -2x+m)=0的三个实根可以作为一个三角形的三边,求实数m的取值范围.2.是否存在质数p,q,使得关于x的一元二次方程p x² -qx+p=0有有理数根.
帮忙解几道一元二次方程的难题
1.如果方程（x-1)( x² -2x+m)=0的三个实根可以作为一个三角形的三边,求实数m的取值范围.
2.是否存在质数p,q,使得关于x的一元二次方程p x² -qx+p=0有有理数根.
已知a,b c都是整数,且a-2b=4,ab+c的平方-1=0,求a+b+c的值.
请麻烦把过程写出来,

帮忙解几道一元二次方程的难题1.如果方程（x-1)( x² -2x+m)=0的三个实根可以作为一个三角形的三边,求实数m的取值范围.2.是否存在质数p,q,使得关于x的一元二次方程p x² -qx+p=0有有理数根.
（一）易知,x=1是原方程的一个根.由题设,方程x²-2x+m=0的两根必为正数,故由求根公式得两根：1+√(1-m),1-√(1-m).(0＜m≤1).∴原方程的三根为0＜1-√(1-m)≤1≤1+√(1-m).(0＜m≤1).再由三角形三边关系得：1+1-√(1-m)＞1+√(1-m).===>√(1-m)＜1/2.===>m＞3/4.∴3/4＜m≤1.(二）易知,⊿=q²-4p²=m²,(m∈Z).一者,4p²=(q+m)(q-m).∴q+m,q-m必为偶数,∴m为奇数,二者,m²=(q+2p)(q-2p).∴q+2p,q-2p为奇数,∴p=2.∴q=5.∴符合题设的数p=2,q=5.(三）∵a-2b=4,ab+c²-1=0.∴a+(-2b)=4,a×(-2b)=2c²-2.∴由韦达定理知,a和-2b是关于x的方程x²-4x+2c²-2=0的两根,⊿=24-8c²=4(6-2c²)≥0.∴0≤c²≤3.且6-2c²是完全平方数,∴c²=1.∴该方程两根为0和4.（1)a=0.-2b=4.c=-1,∴a+b+c=-3,(2)a=0,-2b=4,c=1.∴a+b+c=-1,(3)a=4,-2b=0,c=-1,∴a+b+c=3,(4)a=4,-2b=0,c=1,∴a+b+c=5,综上可知a+b+c=±3,或-1,或5.

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这里有详细解答，希望能帮到你

1) (-9)-(-13)+(-20)+(-2)
(2) 3+13-(-7)/6
(3) (-2)-8-14-13
(4) (-7)*(-1)/7+8
(5) (-11)*4-(-18)/18
(6) 4+(-11)-1/(-3)
(7) (-17)-6-16/(-18)
(8) 5/7+(-1)-(-8)
(9...

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1) (-9)-(-13)+(-20)+(-2)
(2) 3+13-(-7)/6
(3) (-2)-8-14-13
(4) (-7)*(-1)/7+8
(5) (-11)*4-(-18)/18
(6) 4+(-11)-1/(-3)
(7) (-17)-6-16/(-18)
(8) 5/7+(-1)-(-8)
(9) (-1)*(-1)+15+1
(10) 3-(-5)*3/(-15)
(11) 6*(-14)-(-14)+(-13)
(12) (-15)*(-13)-(-17)-(-4)
(13) (-20)/13/(-7)+11
(14) 8+(-1)/7+(-4)
(15) (-13)-(-9)*16*(-12)
(16) (-1)+4*19+(-2)
(17) (-17)*(-9)-20+(-6)
(18) (-5)/12-(-16)*(-15)
(19) (-3)-13*(-5)*13
(20) 5+(-7)+17-10
(21) (-10)-(-16)-13*(-16)
(22) (-14)+4-19-12
(23) 5*13/14/(-10)
(24) 3*1*17/(-10)
(25) 6+(-12)+15-(-15)
(26) 15/9/13+(-7)
(27) 2/(-10)*1-(-8)
(28) 11/(-19)+(-14)-5
(29) 19-16+18/(-11)
(30) (-1)/19+(-5)+1
(31) (-5)+19/10*(-5)
(32) 11/(-17)*(-13)*12
(33) (-8)+(-10)/8*17
(34) 7-(-12)/(-1)+(-12)
(35) 12+12-19+20
(36) (-13)*(-11)*20+(-4)
(37) 17/(-2)-2*(-19)
(38) 1-12*(-16)+(-9)
(39) 13*(-14)-15/20
(40) (-15)*(-13)-6/(-9)
(41) 15*(-1)/12+7
(42) (-13)+(-16)+(-14)-(-6)
(43) 14*12*(-20)*(-13)
(44) 17-9-20+(-10)
(45) 12/(-14)+(-14)+(-2)
(46) (-15)-12/(-17)-(-3)
(47) 6-3/9/(-8)
(48) (-20)*(-15)*10*(-4)
(49) 7/(-2)*(-3)/(-14)
(50) 13/2*18*(-7)
答案：
1 -18
2 103/6
3 -37
4 9
5 -43
6 -(20/3)
7 -(199/9)
8 54/7
9 17
10 2
11 -83
12 216
13 1021/91
14 27/7
15 -1741
16 73
17 127
18 -(2885/12)
19 842
20 5
21 214
22 -41
23 -(13/28)
24 -(51/10)
25 24
26 -(268/39)
27 39/5
28 -(372/19)
29 15/11
30 -(77/19)
31 -(29/2)
32 1716/17
33 -(117/4)
34 -17
35 25
36 2856
37 59/2
38 184
39 -(731/4)
40 587/3
41 23/4
42 -37
43 43680
44 -22
45 -(118/7)
46 -(192/17)
47 145/24
48 -12000
49 -(3/4)
50 -819
①14.59+x-25.31=0
x=10.72
②x-48.32+78.51=80
x=49.81
③820-16x=45.5×8
x=28.5
④(x-6)×7=2x
x=8.4
⑤3x+x=18
x=4.5
⑥0.8+3.2=7.2
x=5
⑦12.5-3x=6.5
x=2
⑧1.2(x-0.64)=0.54
x=1.09
⑨x+12.5=3.5x
x=5
⑩8x-22.8=1.2
x=3
11 x+2=3
x=1
12 x+32=33
x=1
13 x+6=18
x=12
14 4+x=47
x=43
15 19-x=8
x=11
16 98-x=13
x=85
17 66-x=10
x=56
18 5x=10
x=2
19 3x=27
x=9
20 7x=7
x=1
21 8x=8
x=1
22 9x=9
x=1
23 10x=100
x=10
24 66x=660
x=10
25 7x=49
x=7
1、已知A,B是方程x^2+2x-5=0的两个实数根,
求(A^2+2AB+2A)(B^2+2AB+2B)的值.
由A,B是方程x^2+2x-5=0的两个实数根得:
AB=-5,A+B=-2
A^2+2AB+2A)(B^2+2AB+2B)
=AB(A+2B+2)(B+2A+2)
=-5(-2+B+2)(-2+A+2)
=-5AB
=25
2、1/2(x+y+z)方+1/2(x-y-z)(x-y+z)-z(x+y),其中x-y=6,xy=21.要详细步骤
化简得:
1/2(x+y+z)方+1/2(x-y-z)(x-y+z)-z(x+y)=
1/2[(x+y)方+2z(x+y)+z方]+1/2[(x-y)方-z方]-z(x+y)=
1/2(x+y)方+1/2(x-y)方=x方+y方
由x-y=6,xy=21得,x方+y方=(x-y)方+2xy=78
3、a^2-ab+2b^2=3 求2ab-2a^2-4b^2-7的值
2ab-2a^2-4b^2-7
=2(ab-a^2-2b^2)-7
=-2(a^2-ab+2b^2)-7
=(-2)*3-7
=-6-7=-13
4、若A=2x^2+3xy-2x-3,B=-x^2+xy+2,且3A+6B的值与x无关,求y的值
解:
3A+6B=6x^2+9xy-6x-9-6x^2+6xy+12
=15xy-6x+3
=x(15y-6)+3
5、9x+6x^2 -3(x-2/3x^2).其中x=-2
9x+6x2 -3(x-2/3x2)
=9x+6x2-3x+2x2
=8x2+6x
=8×(-2)2+6×(-2)
=32-12
=20
6、1/4(-4x^2+2x-8)-(1/2x-1),其中x=1/2
1/4(-4x2+2x-8)-(1/2x-1)
=-x2+1/2x-2-1/2x+1
=-x2-1
=-(1/2)2-1
=-1/4-1
=-5/4
7、3x'y-[2x'y-(2xyz-x'z)-4x'z]-xyz,其中x=-2,y=-3,z=1,
:3x'y-[2x'y-(2xyz-x'z)-4x'z]-xyz
=3x'y-2x'y+2xyz-x'z+4x'z-xyz
=x'y-xyz+3x'z
=4*(-3)-2*3*1+3*4*1
=-12-6+12
=-6
8、(5a^2-3b^2)+(a^2+b^2)-(5a^2+3b^2),其中a=-1,b=1
=5a^2-3b^2+a^2+b^2-5a^2-3b^2
=a^2-5b^2
=(-1)^2-5*1^2
=1-5
=-4
9、2(a^2b+ab^2)-2(a^2 b-1)-2ab^2 -2其中a=-2,b=2
=2a^2b+2ab^2-2a^2b+2-2ab^2-2
=0
10、(X-2分之1Y-1)(X-2分之1Y+1)-(X-2分之1Y-1)的平方
其中X=1.7,Y=3.9(先化简再求值)
[(X-2分之1Y)-1][(X+2分之1Y)+1]-(X-2分之1Y-1)平方
=(X+2分之1Y)平方-1-(X-2分之1Y)平方+2(X-2分之1Y)-1
=(X+2分之1Y)平方-(X-2分之1Y)平方+2(X-2分之1Y)-2
=2XY+2X-Y-2
=3.9*2.4+1.4
=10.76
11. x（x+2y）-（x+1）²+2x
=x²+2xy-(x²+2x+1)+2x
=2xy-1
12.(5a^2-3b^2)+(a^2+b^2)-(5a^2+3b^2),其中a=-1,b=1
=5a^2-3b^2+a^2+b^2-5a^2-3b^2
=a^2-5b^2
=(-1)^2-5*1^2
=1-5
=-4
13.2(a^2b+ab^2)-2(a^2 b-1)-2ab^2 -2其中a=-2,b=2
=2a^2b+2ab^2-2a^2b+2-2ab^2-2
=0
14. 2/x+3 - 3/3-x - x+15/x平方-9
其中x=根号3 - 2
2/(x+3)- 3/(3-x) - (x+15)/(x^2-9)
= 2/(x+3)+3/(x-3) - (x+15)/(x^2-9)
=(2x-6+3x+9)/(x^2-9) - (x+15)/(x^2-9)
=(5x+3)/(x^2-9) - (x+15)/(x^2-9)
=(5x+3-x-15)/(x^2-9)
=4(x-3)/(x^2-9)
=4/(x+3)
=4/(√3-2+3)
=2√3-2
15.[ 3a/a+1 -a/a-1 ] ÷a/a平方 -1
其中a=2 -根号2
[ 3a/(a+1) -a/(a-1) ] ÷a/(a^2 -1)
=[3a(a-1)-a(a+1)]/(a^2 -1)÷a/(a^2 -1)
=2a(a-2)/(a^2 -1)÷a/(a^2 -1)
=2a(a-2)/(a^2 -1)*(a^2 -1)/a
=2(a-2)
=2(2-√2-2)
=-2√2
16.（a+b)^2=7,(a-b)^2=4,求a^2+b^2和ab的值
（a+b)^2=7
a^2+b^2+2ab=7 (1)
(a-b)^2=4
a^2+b^2-2ab=4 (2)
(1)+(2)
2(a^2+b^2)=11
a^2+b^2=11/2
代入(1)
11/2+2ab=7
ab=3/4
17.[y/(x-y)]+[y^3/(x^3-2x^2y+xy^2)]/[(xy+y^2)/(y^2-x^2)]
x=2/3,y=-3
[y/(x-y)]+[y^3/(x^3-2x^2y+xy^2)]/[(xy+y^2)/(y^2-x^2)]
=[y/(x-y)]+[y^3/x(x-y)^2]/[y(x+y)/(y-x)(y+x)]
=[y/(x-y)]+[y^3/x(x-y)^2]*[(y-x)/y]
=[y/(x-y)]+[y^2/x(y-x)]
=[xy/x(x-y)]-[y^2/x(x-y)]
=(xy-y^2)/x(x-y)
=y(x-y)/x(x-y)
=y/x
=-3/(2/3)
=-3*(3/2)
=-9/2
18.〔（3x-2分之1的y）的2次方+3y（x-12分之y）〕除以〔（2x+y）的2次方-4y（x+4分之1的y）〕，其中x=7.8，
[(3X-Y/2)^2+3Y(X-Y/12)]/[(2X+Y)^2-4Y(X+Y/4)]
=(9X^2-3XY-Y^2/4+3XY-Y^2/4)/(4X^2+4XY+Y^2-4XY-Y^2)
=(9X^2)/(4X^2)
=9/4
19.已知a+b=-5,2a-b=-1,求ab(b+b^2）-b^2(ab-a)+2a(a-b^2)
原式=ab²(b+1)-ab²(b-1)+2a(a-b²)
=ab²[b+1-(b-1)]+2a(a-b²)
=2ab²+2a²-2ab²
=2a²
∵a+b=-5,2a-b=-1
∴a=-2
∴原式=2a²
=2（-2）²
=8
20.（-y）.（y+2x）-（2y+x）.（2y-x）当x=1，y=2时
（-y）.（y+2x）-（2y+x）.（2y-x）
=(x-y-x)(x+y+x)-4y^2+x^2
=x^2-(y+x)^2-4y^2+x^2
=2-9-16=-23
21.（a+2b）.（a-2b）-2（2a+b）.（2a-b）当a=1，b=2时
原式
=a^2-4b^2-8a^2+2b^2
=-7a^2-2b^2
=-7-8=-15
22.（x+2y)(x-2y)-(2x-y)(-2-y),其中x=8,y=-8
x+2y)(x-2y)-(2x-y)(-2-y)
=x^2-4y^2-(-4x-2xy+2y+y^2)
=x^2-4y^2+4x+2xy-2y-y^2
=x^2-5y^2+4x+2xy-2y
当x=8,y=8时
x^2-5y^2+4x+2xy-2y=64-5*64+32+2*64-16
=64*（1-5+2）+32-16
=-128+16=-112
23.(1-2a)(1+2a)(1+4a平方）（1+16a四次方），其中a=-二分之一
2.(1-2a)(1+2a)(1+4a^2)(1+16a^4)
=(1-4a^2)(1+4a^2)(1+16a^4)
=(1-16a^4)(1+16a^4)
=1-256a^8
当a=1/2时
1-256a^8=1-256*(1/2)^8=1-1=0
24.3x^2(x^2-x-1)-x(2x^2-2x-3),其中x=1/2
3x^2(x^2-x-1)-x(2x^2-2x-3)
=3x^4-3x^3-3x^2-2x^3+2x^2+3x
=3x^4-5x^3-x^2+3x
=3*(1/2)^4-5*(1/2)^3-(1/2)^2+3*(1/2)
=3/16-5/8-1/4+3/2
=13/16
25.（3x+2)(3x-2)-5x(x-1)-(2x-1)^2其中x=-1/4
(3x+2)(3x-2)-5x(x-1)-(2x-1)^2
=9x^2-4-5x^2+5x-4x^2+4x-1
=9x-5
=9*(-1/4)-5
=-9/4-20/4
=-29/4
终于完成了

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拜托第一个是三次方程好不好 . 这算个中等题,几年没算了,在手机上不好表达.
不想请教老师的话,问同学:你也肯定能算出的,把题多读几遍 善于发现隐藏条件！