In the first year of mathematics, the addition and subtraction of fractions in the second lesson, th

Solution:1Fill a vacancy.

1) The common denominator of fractions 3 a and 2 5a is (5a).

2) The common denominator of fractions 1 a 2b and 1 ab 2 is (a 2b 2).

2.Compute. 1）(1/a)+(1/b)=(b/ab)+(a/ab)

a+b)/(ab).

2）(1/3x)--1/2x)=(2/6x)--3/6x)

1/6x。3）(b/a)+(b/4a^2)=(4ab/4a^2)+(b/4a^2)

4ab+b)/(4a^2)。

4）(2/m--n)+(2/m+n)=2(m+n)/(m--n)(m+n)+2(m--n)/(m+n)(m--n)

2m+2n)/(m^2--n^2)+(2m--2n)/(m^2--n^2)

4m)/(m^2--n^2).

5）[3/(x--2)(x+1)](1/x--2)=3/(x--2)^2(x+1).

6）(x/x+1)--x^2+1)/(x^2+x)=(x^2/x^2+x)--x^2+1/x^2+x)

1/(x^2+x)。

7）(2/a^2--1)+(1/a+1)=(2/a^2--1)+(a--1/a^2--1)

a+1)/(a^2--1)

1/(a--1).

8）(m+15)/(m^2--9)--3/(m--3)=(m+15)/(m^2--9)--3(m+3)/(m^2--9)

m+15)--3m+9)]/(m^2--9)

-2m+6)/(m^2--9)

2(m--3)/(m+3)(m--3)

2/(m+3)。

I kept typing, and then I found out that I had already gotten it, so I silently deleted it.

A 6a will know one below you don't write clearly don't come to my answer call me **send me I tell you.

Solution:1Fill a vacancy.

1) The common denominator of fractions 3 a and 2 5a is (5a).

2) The common denominator of fractions 1 a 2b and 1 ab 2 is (a 2b 2).

2.Compute. 1）(1/a)+(1/b)=(b/ab)+(a/ab)

a+b)/(ab).

2）(1/3x)--1/2x)=(2/6x)--3/6x)

1/6x。3）(b/a)+(b/4a^2)=(4ab/4a^2)+(b/4a^2)

4ab+b)/(4a^2)。

4）(2/m--n)+(2/m+n)=2(m+n)/(m--n)(m+n)+2(m--n)/(m+n)(m--n)

2m+2n)/(m^2--n^2)+(2m--2n)/(m^2--n^2)

4m)/(m^2--n^2).

5）[3/(x--2)(x+1)](1/x--2)=3/(x--2)^2(x+1).

6）(x/x+1)--x^2+1)/(x^2+x)=(x^2/x^2+x)--x^2+1/x^2+x)

1/(x^2+x)。

7）(2/a^2--1)+(1/a+1)=(2/a^2--1)+(a--1/a^2--1)

a+1)/(a^2--1)

1/(a--1).

8）(m+15)/(m^2--9)--3/(m--3)=(m+15)/(m^2--9)--3(m+3)/(m^2--9)

m+15)--3m+9)]/(m^2--9)

-2m+6)/(m^2--9)

2(m--3)/(m+3)(m--3)

2/(m+3)。

Have you done it yourself first, whether it's right or wrong.

If you don't have time today, I'll do it for you tomorrow.

Solution: From the known, b+c=-a,c+a=-b,a+b=-c,abc≠0, a≠0,b≠0,c≠0

Original formula =-a-b-c=- a+b+c =0

Satisfied, insist on your learning and progress.