1:m-1/x + n-m/x
2:1/m +3/m -5/m
3:a²/a+b +b²+2ab/a+b
4:x-2y/2x-y - x+y/2x-y
5:m+2n/n-m + n/n-m -2n/n-m
6:x²-5/x-2 - x/x-2 -1+x/2-x
7:先化简,再求值。
x²+1/x+1 - 2/x+1 。其中x=1/100
求解答过程!
谢谢!
第1个回答 推荐于2017-12-15
2 a+1-a+3 a2-4a-5÷a2-9 a2-3a-10.
解 原式=[x+2 x(x-2)-x-1(x-2)2]�6�1x 4-x (括号内分式的分母中的多项式式分解因式.分式的除法法则)
=[(x+2)(x-2)x(x-2)2-x(x-1)x(x-2)2]�6�1x4-x(异分母的分式减法的法则)
=x2-4-x2+x x(x-2)2�6�1x4-x (整式运算)
=x-4x(x-2)2�6�1x4-x (合并同类项)
=x-4 x(x-2)2�6�1(-xx-4) (分式的符号法则)
=-1(x-2)2. (分式的乘法法则)
计算x+y x2-xy +(x2-y2 x)2�6�1(1 y-x)3.
解 原式=x+y x(x-y)+(x+y)2(x-y)2x2�6�11(y-x)3
=x+y x(x-y)-(x+y)2 x2(x-y)
=x2+xy-x2-2xy-y2 x2(x-y)
=-xy-y2 x2(x-y)=-xy+y2 x2(x-y).
x-y+4xy x-y)(x+y-4xyx+y)
答案x2-y2
[1 (a+b)2-1(a-b)2]÷(1a+b-1a-b)
答案2a (a+b)(a-b);
x x-y�6�1 y2 x+y-x4y x4-y4÷x2 x2+y2
答案-xy x+y
3x-2 x2-x-2+(1-1x+1)÷(1+1x-1)
答案x2 (x+1)(x-2);
(2x x+1+2 x-1+4x x2-1)×(2x x+1+2 x-1-4x x2-1).
答案4
(2m^2-4m)/(2-m)(m-1)-(1+m)/(1-m^2)
=2m(m-2)/(2-m)(m-1)-(1+m)/(1-m)(1+m)
=-2m/(m-1)-1/(1-m)
=(2m-1)/(1-m)
(-1)-a^2)/(a-1)-a
=(1-a-a^2-a^2+a)/(a-1)
=-(2a^2-1)/(a-1)
(5/x-1)-(3/x+2)+(3/x+3)-(5/x-2)
原式=[5(x+2)-3(x-1)]/(x-1)(x+2)-[5(x+3)-3(x-2)]/(x-2)(x+3)
=(2x+13)/(x�0�5+x-2)-(2x+21)/(x�0�5+x-6)
=[(2x+13)(x�0�5+x-6)-(2x+21)(x�0�5+x-2)]/(x�0�5+x-6)(x�0�5+x-2)
=(2x�0�6+16x�0�5+x-78-2x�0�6-23x�0�5-17x+42)/(x�0�5+x-6)(x�0�5+x-2)
=(-7x�0�5-16x-36)/(x^4+2x�0�6-7x�0�5-8x+12)
1/6x-4y - 1/6x+4y +3x/4y^2-9x^2
=[6x+4y-(6x-4y)-12x]/(36x^2-16y^2)
=(8y-12x)/(36x^2-16y^2)
=4(2y-3x)/[4(3x+2y)(3x-2y)]
=-1/(3x+2y)
(2)1/1-x +1/1+x +2/1+x^2 +4/1+x^4
=2/(1-x^2)+2/(1+x^2)+4/(1+x^4)
=4/(1-x^4)+4/(1+x^4)
=8/(1-x^8)
1.(2x分之3) + 2 = 0
2.(x-1分之x) + (x+1分之2) =1
3.(x+1分之1) - (x�0�5+3x+2分之x�0�5)=-1
3/2x=-2
3=-4x
x=-3/4
x/(x-1)+2/(x+1)=1
x(x+1)+2(x-1)=(x-1)(x+1)
x^2+x+2x-2=x^2-1
3x=1
x=1/3
1/(x+1)-x^2/(x^2+3x+2)=-1
1/(x+1)-x^2/(x+1)(x+2)=-1
(x+2)/(x+1)(x+2)-x^2(x+1)(x+2)=-1
x+2-x^2=-(x+1)(x+2)
x^2-x-2=x^2+3x+2
4x=-4
x=-1
拓展:原题=1-1/2+1/2-1/3....+1/99-1/100
=1-1/100
(2):根据(1)得:
1-1/2+1/2-1/3+.....+1/N-1/(N+1)
=1-1/(N+1)
3.(1)1/1*2+1/2*3+1/3*4+...+1/99*100=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/99-1/100)=1-1/100=99/100
(2)1/1*2+1/2*3+1/3*4+...+1/n(n+1)==(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+[1/n-1/(n+1)]=1-1/(n+1)=n/(n+1)追问
解 原式=[x+2 x(x-2)-x-1(x-2)2]�6�1x 4-x (括号内分式的分母中的多项式式分解因式.分式的除法法则)
=[(x+2)(x-2)x(x-2)2-x(x-1)x(x-2)2]�6�1x4-x(异分母的分式减法的法则)
=x2-4-x2+x x(x-2)2�6�1x4-x (整式运算)
=x-4x(x-2)2�6�1x4-x (合并同类项)
=x-4 x(x-2)2�6�1(-xx-4) (分式的符号法则)
=-1(x-2)2. (分式的乘法法则)
计算x+y x2-xy +(x2-y2 x)2�6�1(1 y-x)3.
解 原式=x+y x(x-y)+(x+y)2(x-y)2x2�6�11(y-x)3
=x+y x(x-y)-(x+y)2 x2(x-y)
=x2+xy-x2-2xy-y2 x2(x-y)
=-xy-y2 x2(x-y)=-xy+y2 x2(x-y).
x-y+4xy x-y)(x+y-4xyx+y)
答案x2-y2
[1 (a+b)2-1(a-b)2]÷(1a+b-1a-b)
答案2a (a+b)(a-b);
x x-y�6�1 y2 x+y-x4y x4-y4÷x2 x2+y2
答案-xy x+y
3x-2 x2-x-2+(1-1x+1)÷(1+1x-1)
答案x2 (x+1)(x-2);
(2x x+1+2 x-1+4x x2-1)×(2x x+1+2 x-1-4x x2-1).
答案4
(2m^2-4m)/(2-m)(m-1)-(1+m)/(1-m^2)
=2m(m-2)/(2-m)(m-1)-(1+m)/(1-m)(1+m)
=-2m/(m-1)-1/(1-m)
=(2m-1)/(1-m)
(-1)-a^2)/(a-1)-a
=(1-a-a^2-a^2+a)/(a-1)
=-(2a^2-1)/(a-1)
(5/x-1)-(3/x+2)+(3/x+3)-(5/x-2)
原式=[5(x+2)-3(x-1)]/(x-1)(x+2)-[5(x+3)-3(x-2)]/(x-2)(x+3)
=(2x+13)/(x�0�5+x-2)-(2x+21)/(x�0�5+x-6)
=[(2x+13)(x�0�5+x-6)-(2x+21)(x�0�5+x-2)]/(x�0�5+x-6)(x�0�5+x-2)
=(2x�0�6+16x�0�5+x-78-2x�0�6-23x�0�5-17x+42)/(x�0�5+x-6)(x�0�5+x-2)
=(-7x�0�5-16x-36)/(x^4+2x�0�6-7x�0�5-8x+12)
1/6x-4y - 1/6x+4y +3x/4y^2-9x^2
=[6x+4y-(6x-4y)-12x]/(36x^2-16y^2)
=(8y-12x)/(36x^2-16y^2)
=4(2y-3x)/[4(3x+2y)(3x-2y)]
=-1/(3x+2y)
(2)1/1-x +1/1+x +2/1+x^2 +4/1+x^4
=2/(1-x^2)+2/(1+x^2)+4/(1+x^4)
=4/(1-x^4)+4/(1+x^4)
=8/(1-x^8)
1.(2x分之3) + 2 = 0
2.(x-1分之x) + (x+1分之2) =1
3.(x+1分之1) - (x�0�5+3x+2分之x�0�5)=-1
3/2x=-2
3=-4x
x=-3/4
x/(x-1)+2/(x+1)=1
x(x+1)+2(x-1)=(x-1)(x+1)
x^2+x+2x-2=x^2-1
3x=1
x=1/3
1/(x+1)-x^2/(x^2+3x+2)=-1
1/(x+1)-x^2/(x+1)(x+2)=-1
(x+2)/(x+1)(x+2)-x^2(x+1)(x+2)=-1
x+2-x^2=-(x+1)(x+2)
x^2-x-2=x^2+3x+2
4x=-4
x=-1
拓展:原题=1-1/2+1/2-1/3....+1/99-1/100
=1-1/100
(2):根据(1)得:
1-1/2+1/2-1/3+.....+1/N-1/(N+1)
=1-1/(N+1)
3.(1)1/1*2+1/2*3+1/3*4+...+1/99*100=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/99-1/100)=1-1/100=99/100
(2)1/1*2+1/2*3+1/3*4+...+1/n(n+1)==(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+[1/n-1/(n+1)]=1-1/(n+1)=n/(n+1)追问
啥玩意·-·
这是我的题的答案吗·-·