Math, asked by Anonymous, 8 months ago

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Answers

Answered by Anonymous
54

Solution:

\sf{\dfrac{1}{x+a}+\dfrac{1}{x+b}+\dfrac{1}{x+c}+\dfrac{ax}{x^{3}+ax^{2}}+\dfrac{bx}{x^{3}+bx^{2}}+\dfrac{cx}{x^{3}+cx^{2}}}

\sf{\leadsto{\dfrac{1}{x+a}+\dfrac{1}{x+b}+\dfrac{1}{x+c}+\dfrac{ax}{x(x^{2}+ax)}+\dfrac{bx}{x(x^{2}+bx)}+\dfrac{cx}{x(x^{2}+cx)}}}

\sf{\leadsto{\dfrac{1}{x+a}+\dfrac{1}{x+b}+\dfrac{1}{x+c}+\dfrac{a}{x^{2}+ax}+\dfrac{b}{x^{2}+bx}+\dfrac{c}{x^{2}+cx}}}

\sf{\leadsto{\dfrac{1}{x+a}+\dfrac{a}{x^{2}+ax}+\dfrac{1}{x+b}+\dfrac{b}{x^{2}+bx}+\dfrac{1}{x+c}+\dfrac{c}{x^{2}+cx}}}

\sf{Multiply \ denominator \ and \ numerator \ of \ \dfrac{1}{x+a}, }

\sf{\dfrac{1}{x+b}, \ \dfrac{1}{x+c} \ by \  x }

\sf{\leadsto{\dfrac{x}{x^{2}+ax}+\dfrac{a}{x^{2}+ax}+\dfrac{x}{x^{2}+bx}+\dfrac{b}{x^{2}+bx}+\dfrac{x}{x^{2}+cx}+\dfrac{c}{x^{2}+cx}}}

\sf{\leadsto{\dfrac{x+a}{x^{2}+ax}+\dfrac{x+b}{x^{2}+bx}+\dfrac{x+c}{x^{2}+cx}}}

\sf{\leadsto{\dfrac{x+a}{x(x+a)}+\dfrac{x+b}{x(x+b)}+\dfrac{x+c}{x(x+c)}}}

\sf{\leadsto{\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}}}

\sf{\leadsto{\dfrac{1+1+1}{x}}}

\sf{\leadsto{\dfrac{3}{x}}}

\sf\purple{\tt{\therefore{Hence, \ simplified \ form \ is \ \dfrac{3}{x}.}}}

Answered by pulakmath007
78

\displaystyle\huge\red{\underline{\underline{Solution}}}

\frac{1}{x +a }  +  \frac{1}{x +b }  +  \frac{1}{x +c }  + \frac{ax}{  {x}^{3} +a  {x}^{2} }  +  \frac{bx}{ {x}^{3}  +b  {x}^{2} }  +  \frac{cx}{  {x}^{3} +c {x}^{2}  }

 = \frac{1}{x +a }  +  \frac{1}{x +b }  +  \frac{1}{x +c }  + \frac{ax}{  {x}^{2} ( {x} +a ) }  +  \frac{bx}{ {x}^{2} (x +b)  }  +  \frac{cx}{  {x}^{2}(x +c)  }

 = \frac{1}{x +a }  +  \frac{1}{x +b }  +  \frac{1}{x +c }  + \frac{a}{  {x}^{} ( {x} +a ) }  +  \frac{b}{ {x}^{} (x +b)  }  +  \frac{c}{  {x}^{}(x +c)  }

 = \frac{1}{x +a }  +\frac{a}{  {x}^{} ( {x} +a ) } +   \frac{1}{x +b }  +  \frac{b}{ {x}^{} (x +b)  }  + \frac{1}{x +c }     +  \frac{c}{  {x}^{}(x +c)  }

 =   \frac{x + a}{  {x}^{} ( {x} +a ) } +  \frac{x + b}{ {x}^{} (x +b)  }  +  \frac{x + c}{  {x}^{}(x +c)  }

 =  \frac{1}{x}  + \frac{1}{x} + \frac{1}{x}

 =  \frac{3}{x}

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