Which one is correct for the partial fractions of5/((x-1)(3x+2))? 

আংশিক ভগ্নাংশে প্রকাশ:

\(\frac{5}{(x-1)(3x+2)}\)

ধরি,

\(\frac{5}{(x-1)(3x+2)} = \frac{A}{x-1} + \frac{B}{3x+2}\) 🤔

উভয় পক্ষে \((x-1)(3x+2)\) দ্বারা গুণ করে পাই,

\(5 = A(3x+2) + B(x-1)\)

এখন, \(x = 1\) বসালে,

\(5 = A(3(1)+2) + B(1-1)\)

\(5 = 5A + 0\)

\(A = 1\) 🎉

আবার, \(x = -\frac{2}{3}\) বসালে,

\(5 = A(3(-\frac{2}{3})+2) + B(-\frac{2}{3}-1)\)

\(5 = A(0) + B(-\frac{5}{3})\)

\(5 = -\frac{5}{3}B\)

\(B = -3\) 🥳

সুতরাং,

\(\frac{5}{(x-1)(3x+2)} = \frac{1}{x-1} + \frac{-3}{3x+2}\)

\(\frac{5}{(x-1)(3x+2)} = \frac{1}{x-1} - \frac{3}{3x+2}\) 🤩

অতএব, উত্তর:

\(\frac{1}{x-1} - \frac{3}{3x+2}\)