Math, asked by yashpadyal5661, 3 months ago

Resolve Into partial fraction of 2x, +3
x2 (x-1)​

Answers

Answered by henniejieun288
3

Step-by-step explanation:

Answer

Resolve into partial fractions

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=1+

x(x−1)(2x+3)

2x−3

Now assume,

x(x−1)(2x+3)

2x−3

=

x

A

+

x−1

B

+

2x+3

C

After simplifying this, we get

x(x−1)(2x+3)

2x−3

=

x(x−1)(2x+3)

A(x−1)(2x+3)+Bx(2x+3)+Cx(x−1)

A(x−1)(2x+3)+Bx(2x+3)+Cx(x−1)=2x−3

Equating the coefficients, we get

2A+2B+C=0,A+3B−C=0 and −3A=−3

On Solving for A,B and C , we get

A=1,B=

5

−1

,C=

5

−8

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=1+

x

1

5(x−1)

1

5(2x+3)

8

Answered by rajendiransanjay
1

Answer:

mark me as brilliant

Step-by-step explanation:

Answer

Resolve into partial fractions

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=1+

x(x−1)(2x+3)

2x−3

Now assume,

x(x−1)(2x+3)

2x−3

=

x

A

+

x−1

B

+

2x+3

C

After simplifying this, we get

x(x−1)(2x+3)

2x−3

=

x(x−1)(2x+3)

A(x−1)(2x+3)+Bx(2x+3)+Cx(x−1)

A(x−1)(2x+3)+Bx(2x+3)+Cx(x−1)=2x−3

Equating the coefficients, we get

2A+2B+C=0,A+3B−C=0 and −3A=−3

On Solving for A,B and C , we get

A=1,B=

5

−1

,C=

5

−8

x(x−1)(2x+3)

2x

3

+x

2

−x−3

=1+

x

1

5(x−1)

1

5(2x+3)

8

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