帮忙解决几道一元二次方程的题目！
\u8bf7\u5927\u5bb6\u5e2e\u5fd9\u89e3\u51b3\u51e0\u9053\u5173\u4e8e\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0b\u7684\u9898\u76ee\uff01\uff01\u6025\uff011.(a+b\uff09x^2-2ax+(a-b)=0
1 -1
a+b -a+b
(x-1)[(a+b)x-a+b]=0
x=1 x=(a-b)/(a+b)
2\u3001\u5df2\u77e5x1\u3001x2\u662f\u5173\u4e8ex\u7684\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0bx^2+(m+1)x+m+6=0\u7684\u4e24\u5b9e\u6570\u6839\uff0c\u4e14x1^2+x2^2=5,\u6c42m\u7684\u503c\u662f\u591a\u5c11\uff1f
x1+x2=-m-1
x1x2=m+6
x1^2+x2^2
=(x1+x2)^2-2x1x2
=M^2+2m+1-2m-12
=m^2-11=5
m=\u00b14
3.x1+x2=5/2
x1x2=1/2
1/x1+1/x2=(x1+x2)/x1x2=5
1/x1*1/x2=2
x^2-5x+2=0
4\u3001\u82e5x1,x2\u662f\u65b9\u7a0bx^2-x=5\u7684\u4e24\u6839\uff0c\u5219x1^2+x2^2=\uff1f
x1+x2=1
x1x2=-5
x1^2+x2^2
=(x1+x2)^2-2x1x2
=11
\u9898\u4e00\uff1a\u8bbe\u6e38\u73a9\u7684\u540c\u5b66\u4e2d\u6709x\u4f4d\u7537\u540c\u5b66,\u56e0\u4e3a\u6bcf\u4e2a\u4eba\u770b\u4e0d\u89c1\u81ea\u5df1\u7684\u5e3d\u5b50\uff0c\u6839\u636e\u4e00\u4f4d\u7537\u540c\u5b66\u8bf4\u201c\u6211\u770b\u89c1\u9ec4\u5e3d\u5b50\u6570\u76ee\u4e0e\u7ea2\u5e3d\u5b50\u6570\u76ee\u76f8\u7b49\u3002\u201d\u53ef\u77e5\u7537\u540c\u5b66\u6bd4\u5973\u540c\u5b66\u591a\u4e00\u4eba\uff0c\u6e38\u73a9\u7684\u540c\u5b66\u4e2d\u6709x-1\u4f4d\u5973\u540c\u5b66\uff0c\u6839\u636e\u4e00\u4f4d\u5973\u540c\u5b66\u8bf4\u201c\u6211\u770b\u89c1\u9ec4\u5e3d\u5b50\u6570\u76ee\u662f\u7ea2\u5e3d\u5b50\u6570\u76ee\u7684\u5e73\u65b9\u201d\u5217\u65b9\u7a0b\u5f97\uff1a
\uff08x-1-1\uff09²=x
x²-4x+4=x
x²-5x+4=0
x=1\u62164
x=1\uff0c\u5219x-1=0\u4e0d\u7b26\u5408\u5b9e\u9645\uff0c\u820d\u53bb
\u2234x=4\uff0cx-1=3\uff0c\u6e38\u73a9\u7684\u540c\u5b66\u4e2d\u5171\u67094\u4f4d\u7537\u540c\u5b66\uff0c3\u4f4d\u5973\u540c\u5b66
\u9898\u4e8c\uff1a\u8bbe\u8fd9\u4e2a\u4e24\u4f4d\u6570\u4e2a\u4f4d\u6570\u5b57\u4e3ax,\u5341\u4f4d\u6570\u5b57\u4e3ax-2,\u5217\u65b9\u7a0b\u5f97\uff1a
(x-2)\u00d7x\u00d73=10(x-2)+x
3x²-6x=11x-20
3x²-17x+20=0
(3x-5)(x-4)=0
x=5/3\u62164
x=5/3\uff0c\u975e\u6574\u6570\u4e0d\u7b26\u5408\u5b9e\u9645\uff0c\u820d\u53bb
\u2234x=4\uff0c\u8fd9\u4e2a\u4e24\u4f4d\u6570\u662f24
x1=(√5+2)/2, x2=(√5-2)/2
2) x^2-x=0
3) 4-4*6*a>=0
a<=1/6
x1+x2=-1/3，x1*x2=a/6
|x1-x2|=2
(x1-x2)^2=4
(x1+x2)^2-4x1*x2=4
1/9-4a/6=4
a=-35/6
4) b^2-4c>=0
b^2>=4c
(b,c)=(2,1)(3,1)(3,2)(4,1)(4,2)(4,3)(4,4)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
1+2+4+6+6=19
5) (1) 6x^2-3x-2=0两根为a，b
a+b=1/2，ab=-1/3
x1+x2=-1/a-1/b=-(a+b)/(ab)=3/2
x1*x2=(-1/a)(-1/b)=1/(ab)=-3
方程：x^2-3/2*x-3=0即 2x^2-3x-6=0
(2) 4x^2-3x-2=0两根为a，b
a+b=3/4，ab=-1/2
x1+x2=a^2+b^2=(a+b)^2-2ab=9/16+1=25/16
x1*x2=a^2*b^2=(ab)^2=1/4
方程：x^2-25/16*x+1/4=0即 16x^2-25x+4=0
6) x1+x2=(2m-1)/m，x1*x2=(m-2)/m
(x1-3)(x2-3)=5m
x1*x2-3(x1+x2)+9=5m
(m-2)/m-3(2m-1)/m+9=5m
(m-2)-3(2m-1)+9m=5m^2
5m^2-4m-1=0
(m-1)(5m+1)=0
m=1或m=-1/5(舍去)
7) 4(m+1)^2-4(4m+1)>=0 (1)
2(m+1)<0 (2)
4m+1<0 (3)
解得 m<-1
（累死了，以后少弄点题吧，遇上我这样的好心人算你的福气。希望能帮到你。祝学习进步！）
1、根据韦达定理[x^2+(x1+x2)x+x1x2=0]写出一元二次方程：x^2+5x+1/4=0 利用公式法解方程得到两数。
2.根据韦达定理[(x1+x2=-b/a) x1x2=c/a]得到 0<-b/a<2 -1< c/a<1 根据[x^2+b/ax+c/a=0]写出一元二次方程：如x^2-x=0
3.a=6 b=2 c=a 因为x1-x2=2 x1+x2=-1/3 x1x2=a/6
所以(x1-x2)^2=x1^2-2x1x2+x2^2=x1^2+2x1x2+x2^2-4x1x2=(x1+x2 )^2-4x1x2=4
1/9+4*a/6=4 解得 a=35/6
4、19个
5、x^2+3/2x+根号2/2=0
6、a=m b=-(2m-1) c=m-2 x1+x2=(2m-1)/m x1x2=(m-2)/m
（x1-3）*（x2-3）=x1x2-3(x1+x2)+9 =5m 代入解出m
7、x1+x2=2(m+1) x1x2=4m+1 2(m+1) <0 得m<-1 4m+1<0得m<-1/4 最后得m<-1
先给1、2两题
直接请教老师去，网 上的不靠谱
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