সমাধান করো :
\(\cfrac{1}{x-a+b}=\cfrac{1}{x}-\cfrac{1}{a}+\cfrac{1}{b}\) Madhyamik 2008


\(\cfrac{1}{x-a+b}=\cfrac{1}{x}-\cfrac{1}{a}+\cfrac{1}{b}\)
বা, \(\cfrac{1}{x-a+b}-\cfrac{1}{x}=-\cfrac{1}{a}+\cfrac{1}{b}\)
বা, \(\cfrac{x-(x-a+b)}{x(x-a+b)} =\cfrac{-b+a}{ab} \)
বা, \(\cfrac{x-x+a-b}{x(x-a+b)} =\cfrac{(a-b)}{ab} \)
বা, \(\cfrac{(a-b)}{x^2-ax+bx}=\cfrac{(a-b)}{ab} \)
বা, \(\cfrac{1}{x^2-ax+bx}=\cfrac{1}{ab} \)
বা, \(x^2-ax+bx=ab \)
বা, \(x^2-ax+bx-ab=0 \)
বা, \(x(x-a)+b(x-a)=0 \)
বা, \((x-a)(x+b)=0 \)

অর্থাৎ,হয় \((x-a)=0 \therefore x=a \)
নয়, \((x+b)=0 \therefore x=-b \)
\(∴x=a\) ও \(x=-b\) হল \( \cfrac{1}{x-a+b}=\cfrac{1}{x}\) \(-\cfrac{1}{a}+\cfrac{1}{b}\) দ্বিঘাত সমীকরনের সমাধান ।

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