1、已知abc≠0,且a+b+c=0,则代数式a^2/bc+b^2/ac+c^2/ab的值为_____2、关于x的方程x+2/x=c+2/c的两个解是x1=c,x2=2/c,则关于x的方程x+2/(x-1)=a+2/(a-1)的两个解是_____
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/27 05:03:42
![1、已知abc≠0,且a+b+c=0,则代数式a^2/bc+b^2/ac+c^2/ab的值为_____2、关于x的方程x+2/x=c+2/c的两个解是x1=c,x2=2/c,则关于x的方程x+2/(x-1)=a+2/(a-1)的两个解是_____](/uploads/image/z/2106314-26-4.jpg?t=1%E3%80%81%E5%B7%B2%E7%9F%A5abc%E2%89%A00%2C%E4%B8%94a%2Bb%2Bc%3D0%2C%E5%88%99%E4%BB%A3%E6%95%B0%E5%BC%8Fa%5E2%2Fbc%2Bb%5E2%2Fac%2Bc%5E2%2Fab%E7%9A%84%E5%80%BC%E4%B8%BA_____2%E3%80%81%E5%85%B3%E4%BA%8Ex%E7%9A%84%E6%96%B9%E7%A8%8Bx%2B2%2Fx%3Dc%2B2%2Fc%E7%9A%84%E4%B8%A4%E4%B8%AA%E8%A7%A3%E6%98%AFx1%3Dc%2Cx2%3D2%2Fc%2C%E5%88%99%E5%85%B3%E4%BA%8Ex%E7%9A%84%E6%96%B9%E7%A8%8Bx%2B2%2F%28x-1%29%3Da%2B2%2F%28a-1%29%E7%9A%84%E4%B8%A4%E4%B8%AA%E8%A7%A3%E6%98%AF_____)
]2tQZo7M="1WԺ%t B95ŏ.9kQbjTDpdr\M}6$g÷%q|4;>k1}Q9iu9SV2.Un 3)xQe ZUT?X ħ{0ATw*ͤzL|)q`8ᷱZv%t:k@(\ 7튳A&%^fuc~("
1、已知abc≠0,且a+b+c=0,则代数式a^2/bc+b^2/ac+c^2/ab的值为_____2、关于x的方程x+2/x=c+2/c的两个解是x1=c,x2=2/c,则关于x的方程x+2/(x-1)=a+2/(a-1)的两个解是_____
1、已知abc≠0,且a+b+c=0,则代数式a^2/bc+b^2/ac+c^2/ab的值为_____
2、关于x的方程x+2/x=c+2/c的两个解是x1=c,x2=2/c,则关于x的方程x+2/(x-1)=a+2/(a-1)的两个解是_____
1、已知abc≠0,且a+b+c=0,则代数式a^2/bc+b^2/ac+c^2/ab的值为_____2、关于x的方程x+2/x=c+2/c的两个解是x1=c,x2=2/c,则关于x的方程x+2/(x-1)=a+2/(a-1)的两个解是_____
1.
a^2/bc+b^2/ac+c^2/ab=(a^3+b^3+c^3)/abc
=[(a+b)^3-3ab^2-3ba^2+c^3]/abc
=[(-c)^3-3ab^2-3ba^2+c^3]/abc
=-3ab(a+b)/abc
=-3(-c)/c
=3
2
x+2/(x-1)=a+2/(a-1)等价于(x-1)++2/(x-1)=(a-1)+2/(a-1)
设x-1=x`,a-1=c`,x=x`+1
则原式等价于x`+2/x`=c`+2/c`,它的两个解是x1`=c`,x2`=2/c`
所以原式解为x1=c`+1=a-1+1=a,x2=2/c`+1=1+2/(a-1)=(a+1)/(a-1)
1.检举a+b+c=0,a+b=-c
a^2/bc+b^2/ac+c^2/ab
通分
=(a^3+b^3+c^3)/abc
=[(a+b)(a^2-ab+b^2)+c^3]/abc
=[-c(a^2-ab+b^2)+c^3]/abc
={-c[(a+b)^2-3ab]+c^3}/abc
=[-c(c^2-3ab)+c^3]/abc
=(-c^2+3abc+c^3)/abc
=3abc/abc
=3
1.因为a+b+c=0,则有a=-(b+c),则有a^2=[-(b+c)]^2=b^2+c^2+2bc
则a^2/bc=c/b+b/c+2 同理可得原式等于
c/b+b/c+2 + c/a+a/c+2 + b/a+a/b+2 =
(b+c)/a + (a+c)/b +(a+b)/c +6
因为a+b...
全部展开
1.因为a+b+c=0,则有a=-(b+c),则有a^2=[-(b+c)]^2=b^2+c^2+2bc
则a^2/bc=c/b+b/c+2 同理可得原式等于
c/b+b/c+2 + c/a+a/c+2 + b/a+a/b+2 =
(b+c)/a + (a+c)/b +(a+b)/c +6
因为a+b+c=0 所以原式等于(-a)/a + (-b)/b + (-c)/c +6
=(-1)+(-1)+(-1)+6=3
2.很显然有一个解是x=a,另一个。。。恕在下愚钝。。。
收起