Question

reslove 9x+7/ [ x-2] [x+3] into partial fractions

Answer 1

Answer:

(x−1)(x+2)

2

9

=

x−1

A

+

x+2

B

+

(x+2)

2

C

=>9=A(x+2)

2

+B(x−1)(x+2)+C(x−1) -(1)

For x=1,=>9=A(3)

2

+B(0)+C(0)

=>9=9A

=>A=1 - (2)

For x=−2,=>9=A(0)+B(0)+C(−2−1)

=>9=−3C

=>C=−3 -(3)

Now , we have found out value of A and C , to find value of B , put x = any real number .

Let x =0 , then equation (1) reduces to 9=4A−2B−C

=>9=4−2B

=>2=−2B

=>B=−1 -(4)

Therefore from (2),(3),(4) we get

(x−1)(x+2)

2

9

=

x−1

1

−

x+2

1

−

(x+2)

2

3

Was this answer helpful?

32

Answer 2

Answer:

it s is easy for g-10 haha

(×+5),(×-5)

good