If frac{(x+1)}{(2 x-1)(3 x+1)}=frac{A}{(2 x-1 | vedclass

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(C) Given that,$\frac{x+1}{(2 x-1)(3 x+1)}=\frac{A}{(2 x-1)}+\frac{B}{(3 x+1)}$Multiplying both sides by $(2x-1)(3x+1)$,we get:$(x+1) = A(3x+1) + B(2x

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$6$

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(C) Given that,$\frac{x+1}{(2 x-1)(3 x+1)}=\frac{A}{(2 x-1)}+\frac{B}{(3 x+1)}$Multiplying both sides by $(2x-1)(3x+1)$,we get:$(x+1) = A(3x+1) + B(2x-1)$$(x+1) = x(3A+2B) + (A-B)$Comparing the coefficients of $x$ and the constant terms on both sides,we get:$3A + 2B = 1$ ... $(i)$$A - B = 1$ ... $(ii)$From equation $(ii)$,$A = B + 1$. Substituting this into equation $(i)$:$3(B+1) + 2B = 1$$3B + 3 + 2B = 1$$5B = -2 \Rightarrow B = -\frac{2}{5}$Now,$A = -\frac{2}{5} + 1 = \frac{3}{5}$We need to find the value of $16A + 9B$:$16A + 9B = 16\left(\frac{3}{5}\right) + 9\left(-\frac{2}{5}\right)$$= \frac{48}{5} - \frac{18}{5} = \frac{30}{5} = 6$

If $\frac{(x+1)}{(2 x-1)(3 x+1)}=\frac{A}{(2 x-1)}+\frac{B}{(3 x+1)}$,then $16 A+9 B$ is equal to

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