Math, asked by sangitabasumatary432, 2 months ago

Resolve partial fraction​

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Answered by mathdude500
1

\large\underline{\bold{Given \:Question - }}

 \sf \: Resolve  \: in \:  to  \: partial \:  fraction  :  \: \dfrac{x - 1}{ {(x - 3)}^{2} }

\large\underline{\bold{Solution-}}

 \bf \: Let \:  \sf \: \dfrac{x - 1}{ {(x - 3)}^{2} } =  \dfrac{a}{x - 3}  + \dfrac{b}{ {(x - 3)}^{2} }  -  - (1)

On taking LCM, we get

\rm :\longmapsto\:x - 1 = a(x - 3) + b -  - (2)

On substituting 'x = 3' in equation (2), we get

\rm :\longmapsto\:3 - 1 = b

\bf\implies \: b \:  =  \: 2 \:  -  - (3)

On substituting 'x = 0' in equation (2), we get

\rm :\longmapsto\:0 - 1 = a {(0 - 3)} + b

\rm :\longmapsto\: - 1 =  - 3a + 2

\rm :\longmapsto\: - 3a =  - 3

\bf\implies \:a \:  =  \: 1  -  - (4)

On substituting the values of a and b, in equation (1), we get

\bf \: \dfrac{x - 1}{ {(x - 3)}^{2} } =  \dfrac{ 1}{(x - 3)}  + \dfrac{2}{ {(x - 3)}^{2} }

Additional Information :-

Partial Fraction :-

Partial fraction decomposition is the breaking down of a rational expression into simplier parts. It is the opposite of adding rational expressions. When adding two rational expressions, there has to be a common denominator.

\begin{gathered}\boxed{\begin{array}{c|c} \sf factor \: in \: denominator & \sf partial \: fraction \: decomposition \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ax + b & \sf \displaystyle \frac{A}{{ax + b}} \\ \\ \sf  {(ax + b)}^{2}  & \sf \displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}}  \\ \\ \sf  {ax}^{2} + bx + c  & \sf \displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}} \end{array}} \\ \end{gathered}

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