If frac{1}{(3-5 x)(2+3 x)}=frac{A}{3-5 x}+fra | vedclass

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(B) Given,$\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$ Multiplying both sides by $(3-5 x)(2+3 x)$,we get: $1 = A(2+3 x) + B(3-5 x)$ $1 =

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$5: 3$

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(B) Given,$\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$ Multiplying both sides by $(3-5 x)(2+3 x)$,we get: $1 = A(2+3 x) + B(3-5 x)$ $1 = (2A + 3B) + x(3A - 5B)$ Comparing the coefficients of $x$ and the constant terms on both sides: $3A - 5B = 0 \implies 3A = 5B \implies A = \frac{5}{3}B$ $2A + 3B = 1$ Substituting $A = \frac{5}{3}B$ into the second equation: $2(\frac{5}{3}B) + 3B = 1$ $\frac{10}{3}B + 3B = 1$ $\frac{10B + 9B}{3} = 1$ $19B = 3 \implies B = \frac{3}{19}$ Now,$A = \frac{5}{3} 	imes \frac{3}{19} = \frac{5}{19}$ Therefore,$A : B = \frac{5}{19} : \frac{3}{19} = 5 : 3$

If $\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$,then $A : B$ is

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