几道数学题
1、(3y+5)/(3y-6)=(1/2)+(5y-4)/(2y-4)2、4/(x^2-4)+(x+3)/(x-2)3、3/(x)+(6)/(x-1)-(x-4)/x(x...
1、(3y+5)/(3y-6)=(1/2)+(5y-4)/(2y-4)
2、4/(x^2-4)+(x+3)/(x-2)
3、 3/(x)+(6)/(x-1)-(x-4)/ x(x-1) =0
4、1/(x+4)-(1)/(x+7)=1/(x+3)-1/(x+6)
记得要过程啊!亲们!谢谢了! 展开
2、4/(x^2-4)+(x+3)/(x-2)
3、 3/(x)+(6)/(x-1)-(x-4)/ x(x-1) =0
4、1/(x+4)-(1)/(x+7)=1/(x+3)-1/(x+6)
记得要过程啊!亲们!谢谢了! 展开
8个回答
展开全部
1、(3y+5)/(3y-6)=(1/2)+(5y-4)/(2y-4)
(3y+5)/3(y-2)=(1/2)+(5y-4)/2(y-2)
[(3y+5)/3(y-2)-(5y-4)/2(y-2)]=1/2
[2(3y+5)-3(5y-4)]/6(y-2)]=1/2
6y+10-15y+12=3(y-2)
22-9y=3y-6
y=7/4
2、4/(x^2-4)+(x+3)/(x-2)
=[4+(x+3(x+2)]/(x^2-4)
=(x^2+5x+10)/(x^2-4)
3、 3/(x)+(6)/(x-1)-(x-4)/ x(x-1) =0
[3(x-1)+6x-(x-4]/ x(x-1)=0
(3x-1+6x-x+4)/ x(x-1)=0
(8x-3)/ x(x-1)=0
8x-3=0
x=3/8
4、1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)
[1/(x+4)+1/(x+6)]=[1/(x+3)+1/(x+7)]
2(x+5)/(x+4)(x+6)=2(x+5)/(x+3)(x+7)
2(x+5){[1/(x+5)/(x+6)]-1/(x+3)(x+7)}=0
2(x+5){[1/(x^2+10x+30)]-1/(x^2+10x+21)]-)}=0
2(x+5){[(x^2+10x+21)]-(x^2+10x+30)]/[(x^2+10x)^2+51(x^2+10x)+30*21]}=0
-18(x+5)/[(x^2+10x)^2+51(x^2+10x)+30*21]}=0
-18(x+5)=0
x=-5
(3y+5)/3(y-2)=(1/2)+(5y-4)/2(y-2)
[(3y+5)/3(y-2)-(5y-4)/2(y-2)]=1/2
[2(3y+5)-3(5y-4)]/6(y-2)]=1/2
6y+10-15y+12=3(y-2)
22-9y=3y-6
y=7/4
2、4/(x^2-4)+(x+3)/(x-2)
=[4+(x+3(x+2)]/(x^2-4)
=(x^2+5x+10)/(x^2-4)
3、 3/(x)+(6)/(x-1)-(x-4)/ x(x-1) =0
[3(x-1)+6x-(x-4]/ x(x-1)=0
(3x-1+6x-x+4)/ x(x-1)=0
(8x-3)/ x(x-1)=0
8x-3=0
x=3/8
4、1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)
[1/(x+4)+1/(x+6)]=[1/(x+3)+1/(x+7)]
2(x+5)/(x+4)(x+6)=2(x+5)/(x+3)(x+7)
2(x+5){[1/(x+5)/(x+6)]-1/(x+3)(x+7)}=0
2(x+5){[1/(x^2+10x+30)]-1/(x^2+10x+21)]-)}=0
2(x+5){[(x^2+10x+21)]-(x^2+10x+30)]/[(x^2+10x)^2+51(x^2+10x)+30*21]}=0
-18(x+5)/[(x^2+10x)^2+51(x^2+10x)+30*21]}=0
-18(x+5)=0
x=-5
展开全部
(1)两边同乘以6(y-2),得 2(3y+5)=3(y-2)+3(5y-4) 6y+10=18y-18 12y=28 y=7/3(2) 通分: 3/((x+4)(x+7))=3/((x+3)(x+6)) (x+4)(x+7)=(x+3)(x+6) 11x+28=9x+18 2x= -10 x= -5
(x-2)^2-4(x+3)^2=0
[(x-2)+2(x+3)][(x-2)-2(x+3)]=0
(3x+4)(-x-8)=0
x=-4/3 x=-8
亲爱的同学 你好!给你两个建议 1 自己答题 学会了 才最重要 因为考试不能用百度.2 学会用百度,想要答案直接在网页里收 就可以了!如果没有在用百度知道! 祝你考上自己理想的学校 好好学习 加油
!
(x-2)^2-4(x+3)^2=0
[(x-2)+2(x+3)][(x-2)-2(x+3)]=0
(3x+4)(-x-8)=0
x=-4/3 x=-8
亲爱的同学 你好!给你两个建议 1 自己答题 学会了 才最重要 因为考试不能用百度.2 学会用百度,想要答案直接在网页里收 就可以了!如果没有在用百度知道! 祝你考上自己理想的学校 好好学习 加油
!
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
1、解:原式等=(3y-6+11)/(3y-6)=1/2+(2y-4+3y)/(2y-4)
1+11/(3y-6)=1/2+3y/(2y-4)+1
11/{3(y-2)}=1/2+3y/{2(y-2)}
两边同时乘以(y-2)得:2y=14/3
y=7/3
2、原式={4+(x+3)(x+2)}/(x^2-4)
=(x^2+5x+10)/(x^2-4)
3、原式={3(x-1)+6x-(x-4)}/{x(x-1)}=0
( 8x+1)/{x(x-1)}=0
x=-1/8
4、原式=1/(x+4)-1/(x+3)=1/(x+7)-1/(x+6)
-1/{(x+3)(x+4)}= -1/{(x+7)(x+6)}
(x+3)(x+4)= ( x+7)(x+6)
x^2+7x+12=x^2+13x+42
-30=6x
x=-5
1+11/(3y-6)=1/2+3y/(2y-4)+1
11/{3(y-2)}=1/2+3y/{2(y-2)}
两边同时乘以(y-2)得:2y=14/3
y=7/3
2、原式={4+(x+3)(x+2)}/(x^2-4)
=(x^2+5x+10)/(x^2-4)
3、原式={3(x-1)+6x-(x-4)}/{x(x-1)}=0
( 8x+1)/{x(x-1)}=0
x=-1/8
4、原式=1/(x+4)-1/(x+3)=1/(x+7)-1/(x+6)
-1/{(x+3)(x+4)}= -1/{(x+7)(x+6)}
(x+3)(x+4)= ( x+7)(x+6)
x^2+7x+12=x^2+13x+42
-30=6x
x=-5
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
一个一个帮你解答:
1.
(3y+5)/(3y-6)=(1/2)+(5y-4)/(2y-4)
(3y+5)(2y-4)=1/2(3y-6)(2y-4)+(5y-4)(3y-6) 左右两端都乘以(3y-6)(2y-4)
6y^2-2y-20=3y^2-12y+12+15y^2-42y+24
3y^2-13y+14=0
(y-2)(3y-7)=0
得:y=7/3
看懂吗?
1.
(3y+5)/(3y-6)=(1/2)+(5y-4)/(2y-4)
(3y+5)(2y-4)=1/2(3y-6)(2y-4)+(5y-4)(3y-6) 左右两端都乘以(3y-6)(2y-4)
6y^2-2y-20=3y^2-12y+12+15y^2-42y+24
3y^2-13y+14=0
(y-2)(3y-7)=0
得:y=7/3
看懂吗?
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
1、(3y+5)/(3y-6)=(1/2)+(5y-4)/(2y-4)
(3y+5)/3(y-2)=(1/2)+(5y-4)/2(y-2)
[(3y+5)/3(y-2)-(5y-4)/2(y-2)]=1/2
[2(3y+5)-3(5y-4)]/6(y-2)]=1/2
6y+10-15y+12=3(y-2)
22-9y=3y-6
y=7/4
2、4/(x^2-4)+(x+3)/(x-2)
=[4+(x+3(x+2)]/(x^2-4)
=(x^2+5x+10)/(x^2-4)
3、 3/(x)+(6)/(x-1)-(x-4)/ x(x-1) =0
[3(x-1)+6x-(x-4]/ x(x-1)=0
(3x-1+6x-x+4)/ x(x-1)=0
(8x-3)/ x(x-1)=0
8x-3=0
x=3/8
4、1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)
[1/(x+4)+1/(x+6)]=[1/(x+3)+1/(x+7)]
2(x+5)/(x+4)(x+6)=2(x+5)/(x+3)(x+7)
2(x+5){[1/(x+5)/(x+6)]-1/(x+3)(x+7)}=0
2(x+5){[1/(x^2+10x+30)]-1/(x^2+10x+21)]-)}=0
2(x+5){[(x^2+10x+21)]-(x^2+10x+30)]/[(x^2+10x)^2+51(x^2+10x)+30*21]}=0
-18(x+5)/[(x^2+10x)^2+51(x^2+10x)+30*21]}=0
-18(x+5)=0
x=-5
】
(1)两边同乘以6(y-2),得 2(3y+5)=3(y-2)+3(5y-4) 6y+10=18y-18 12y=28 y=7/3(2) 通分: 3/((x+4)(x+7))=3/((x+3)(x+6)) (x+4)(x+7)=(x+3)(x+6) 11x+28=9x+18 2x= -10 x= -5
(x-2)^2-4(x+3)^2=0
[(x-2)+2(x+3)][(x-2)-2(x+3)]=0
(3x+4)(-x-8)=0
x=-4/3 x=-8
亲爱的同学 你好!给你两个建议 1 自己答题 学会了 才最重要 因为考试不能用百度.2 学会用百度,想要答案直接在网页里收 就可以了!如果没有在用百度知道! 祝你考上自己理想的学校 好好学习 加油
!赞同0|评论
9 分钟前qsk4015|五级1、解:原式等=(3y-6+11)/(3y-6)=1/2+(2y-4+3y)/(2y-4)
1+11/(3y-6)=1/2+3y/(2y-4)+1
11/{3(y-2)}=1/2+3y/{2(y-2)}
两边同时乘以(y-2)得:2y=14/3
y=7/3
2、原式={4+(x+3)(x+2)}/(x^2-4)
=(x^2+5x+10)/(x^2-4)
3、原式={3(x-1)+6x-(x-4)}/{x(x-1)}=0
( 8x+1)/{x(x-1)}=0
x=-1/8
4、原式=1/(x+4)-1/(x+3)=1/(x+7)-1/(x+6)
-1/{(x+3)(x+4)}= -1/{(x+7)(x+6)}
(x+3)(x+4)= ( x+7)(x+6)
x^2+7x+12=x^2+13x+42
-30=6x
x=-5
(3y+5)/3(y-2)=(1/2)+(5y-4)/2(y-2)
[(3y+5)/3(y-2)-(5y-4)/2(y-2)]=1/2
[2(3y+5)-3(5y-4)]/6(y-2)]=1/2
6y+10-15y+12=3(y-2)
22-9y=3y-6
y=7/4
2、4/(x^2-4)+(x+3)/(x-2)
=[4+(x+3(x+2)]/(x^2-4)
=(x^2+5x+10)/(x^2-4)
3、 3/(x)+(6)/(x-1)-(x-4)/ x(x-1) =0
[3(x-1)+6x-(x-4]/ x(x-1)=0
(3x-1+6x-x+4)/ x(x-1)=0
(8x-3)/ x(x-1)=0
8x-3=0
x=3/8
4、1/(x+4)-1/(x+7)=1/(x+3)-1/(x+6)
[1/(x+4)+1/(x+6)]=[1/(x+3)+1/(x+7)]
2(x+5)/(x+4)(x+6)=2(x+5)/(x+3)(x+7)
2(x+5){[1/(x+5)/(x+6)]-1/(x+3)(x+7)}=0
2(x+5){[1/(x^2+10x+30)]-1/(x^2+10x+21)]-)}=0
2(x+5){[(x^2+10x+21)]-(x^2+10x+30)]/[(x^2+10x)^2+51(x^2+10x)+30*21]}=0
-18(x+5)/[(x^2+10x)^2+51(x^2+10x)+30*21]}=0
-18(x+5)=0
x=-5
】
(1)两边同乘以6(y-2),得 2(3y+5)=3(y-2)+3(5y-4) 6y+10=18y-18 12y=28 y=7/3(2) 通分: 3/((x+4)(x+7))=3/((x+3)(x+6)) (x+4)(x+7)=(x+3)(x+6) 11x+28=9x+18 2x= -10 x= -5
(x-2)^2-4(x+3)^2=0
[(x-2)+2(x+3)][(x-2)-2(x+3)]=0
(3x+4)(-x-8)=0
x=-4/3 x=-8
亲爱的同学 你好!给你两个建议 1 自己答题 学会了 才最重要 因为考试不能用百度.2 学会用百度,想要答案直接在网页里收 就可以了!如果没有在用百度知道! 祝你考上自己理想的学校 好好学习 加油
!赞同0|评论
9 分钟前qsk4015|五级1、解:原式等=(3y-6+11)/(3y-6)=1/2+(2y-4+3y)/(2y-4)
1+11/(3y-6)=1/2+3y/(2y-4)+1
11/{3(y-2)}=1/2+3y/{2(y-2)}
两边同时乘以(y-2)得:2y=14/3
y=7/3
2、原式={4+(x+3)(x+2)}/(x^2-4)
=(x^2+5x+10)/(x^2-4)
3、原式={3(x-1)+6x-(x-4)}/{x(x-1)}=0
( 8x+1)/{x(x-1)}=0
x=-1/8
4、原式=1/(x+4)-1/(x+3)=1/(x+7)-1/(x+6)
-1/{(x+3)(x+4)}= -1/{(x+7)(x+6)}
(x+3)(x+4)= ( x+7)(x+6)
x^2+7x+12=x^2+13x+42
-30=6x
x=-5
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
分式的话,学会通分合并同类项。
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询