(a) Write \(f(x)=\dfrac{Ax+B}{2+3x^2}+\dfrac C{2-x}\). Equating numerators gives \(17x^2-7x+16=(Ax+B)(2-x)+C(2+3x^2)\). Thus \(-A+3C=17,\ 2A-B=-7,\ 2B+2C=16\), giving \(A=-2,B=3,C=5\). Therefore \(\boxed{f(x)=\dfrac{-2x+3}{2+3x^2}+\dfrac5{2-x}}\).

(b) \(\dfrac{-2x+3}{2+3x^2}=(\tfrac32-x)(1-\tfrac32x^2+O(x^4))=\tfrac32-x-\tfrac94x^2+\tfrac32x^3+O(x^4)\). Also \(\dfrac5{2-x}=\tfrac52(1+\tfrac x2+\tfrac{x^2}4+\tfrac{x^3}8+O(x^4))\). Adding gives \(\boxed{f(x)=4+\tfrac14x-\tfrac{13}8x^2+\tfrac{29}{16}x^3+O(x^4)}\).

(c) We need \(|3x^2/2|<1\) and \(|x/2|<1\), so \(\boxed{|x|<\sqrt{2/3}}\).

(a) Write \(f(x)=\dfrac A{1-2x}+\dfrac B{x+2}+\dfrac C{(x+2)^2}\). Then \(24x+13=A(x+2)^2+B(1-2x)(x+2)+C(1-2x)\). At \(x=1/2\), \(A=4\); at \(x=-2\), \(C=-7\). The coefficient of \(x^2\) gives \(A-2B=0\), so \(B=2\). Therefore \(\boxed{f(x)=\dfrac4{1-2x}+\dfrac2{x+2}-\dfrac7{(x+2)^2}}\).

(b) Expand each term: \(4(1+2x+4x^2)+\left(1-\tfrac x2+\tfrac{x^2}4\right)-\tfrac74\left(1-x+\tfrac34x^2\right)+O(x^3)\). Collecting terms gives \(\boxed{\tfrac{13}4+\tfrac{37}4x+\tfrac{239}{16}x^2+O(x^3)}\).

(c) The required conditions are \(|2x|<1\) and \(|x/2|<1\). Their intersection is \(\boxed{|x|<1/2}\).

(a) Write \(f(x)=A/(1+2x)+B/(3-x)+C/(3-x)^2\). Multiplying by the denominator, values \(x=-1/2,3\) give \(A=2,C=-3\). Comparing \(x^2\) gives \(A-2B=-2\), so \(B=2\). Hence \(\boxed{f(x)=\dfrac2{1+2x}+\dfrac2{3-x}-\dfrac3{(3-x)^2}}\).

(b) The expansions are \(2(1-2x+4x^2)+\tfrac23(1+\tfrac x3+\tfrac{x^2}9)-\tfrac13(1+\tfrac{2x}3+\tfrac{x^2}3)+O(x^3)\). Collecting gives \(\boxed{f(x)=\tfrac73-4x+\tfrac{215}{27}x^2+O(x^3)}\).

(a) Let \[f(x)=\frac A{1+x}+\frac B{2+x}+\frac C{(2+x)^2}.\] Multiplying by the denominator, \[2x^2+7x+8=A(x+2)^2+B(x+1)(x+2)+C(x+1).\] At \(x=-1\), \(A=3\); at \(x=-2\), \(C=-2\). Comparing the \(x^2\) coefficients gives \(A+B=2\), so \(B=-1\). Thus \[\boxed{f(x)=\frac3{1+x}-\frac1{2+x}-\frac2{(2+x)^2}}.\]

(b) Using the binomial series, \[\frac3{1+x}=3(1-x+x^2+\cdots),\] \[-\frac1{2+x}=-\frac12\left(1-\frac x2+\frac{x^2}4+\cdots\right),\] \[-\frac2{(2+x)^2}=-\frac12\left(1-x+\frac{3x^2}4+\cdots\right).\] Adding the terms gives \[\boxed{f(x)=2-\frac94x+\frac52x^2+\cdots}.\] The combined expansion is valid for \(|x|<1\).

(a) Set \[f(x)=\frac A{x-2}+\frac{Bx+C}{2x^2+3}.\] Equating numerators gives \[5x^2+8x-3=A(2x^2+3)+(Bx+C)(x-2).\] At \(x=2\), \(33=11A\), so \(A=3\). Then comparing coefficients gives \(B=-1\) and \(C=6\). Hence \[\boxed{f(x)=\frac3{x-2}+\frac{-x+6}{2x^2+3}}.\]

(b) Expand the two terms: \[\frac3{x-2}=-\frac32(1-x/2)^{-1}=-\frac32-\frac34x-\frac38x^2+\cdots,\] \[\frac{6-x}{2x^2+3}=\left(2-\frac x3\right)\left(1+\frac23x^2\right)^{-1}=2-\frac x3-\frac43x^2+\cdots.\] Therefore \[\boxed{f(x)=\frac12-\frac{13}{12}x-\frac{41}{24}x^2+\cdots}.\] The combined expansion is valid for \(|x|<\sqrt{3/2}\).

(a) Write \[f(x)=\frac A{x+2}+\frac{B+Cx}{x^2+3}.\] Multiply through: \[14-3x+2x^2=A(x^2+3)+(B+Cx)(x+2).\] Comparing coefficients gives \[A+C=2,\quad B+2C=-3,\quad3A+2B=14.\] Solving yields A=4, B=1, C=−2, so \[\boxed{f(x)=\frac4{x+2}+\frac{1-2x}{x^2+3}}.\]

(b) Expand the two fractions: \[\frac4{x+2}=2(1+x/2)^{-1}=2-x+\frac{x^2}{2}+O(x^3),\] \[\frac{1-2x}{x^2+3}=\frac{1-2x}{3}(1+x^2/3)^{-1}=\frac13-\frac23x-\frac19x^2+O(x^3).\] Adding gives \[\boxed{f(x)=\frac73-\frac53x+\frac7{18}x^2+O(x^3)}.\] These expansions are valid for |x|<√3.

(a) Write \(f=A/(1+2x)+B/(1-2x)+C/(2+x)\). Multiplying through and substituting \(x=-1/2,1/2,-2\), respectively, gives \(A=-2,B=1,C=4\). Therefore \[\boxed{f(x)=-\frac2{1+2x}+\frac1{1-2x}+\frac4{2+x}}.\]

(b) \[\begin{aligned}f(x)&=-2(1-2x+4x^2)+(1+2x+4x^2)+2(1-x/2+x^2/4)+O(x^3)\\&=\boxed{1+5x-\frac72x^2+O(x^3)}.\end{aligned}\] This is valid for \(|x|<1/2\).

(a) Write \[f(x)=\frac A{1-x}+\frac B{2+3x}+\frac C{(2+3x)^2}.\] Multiplying through gives \(8+5x+12x^2=A(2+3x)^2+B(1-x)(2+3x)+C(1-x)\). At \(x=1\), \(A=1\); at \(x=-2/3\), \(C=6\); the constant coefficient then gives \(B=-1\). Thus \[\boxed{f(x)=\frac1{1-x}-\frac1{2+3x}+\frac6{(2+3x)^2}}.\]

(b) \[\begin{aligned}f(x)&=(1+x+x^2)-\frac12\left(1-\frac32x+\frac94x^2\right)+\frac32\left(1-3x+\frac{27}4x^2\right)+O(x^3)\\&=\boxed{2-\frac{11}4x+10x^2+O(x^3)}.\end{aligned}\] The expansion is valid for \(|x|<2/3\).

(a) Write \[f(x)=\frac{A}{3x+1}+\frac{Bx+C}{x^2+4}.\] Then \[36-4x-3x^2=A(x^2+4)+(Bx+C)(3x+1).\] Comparing coefficients gives \(A+3B=-3\), \(B+3C=-4\), \(4A+C=36\). Solving, \(A=9,B=-4,C=0\). Hence \[\boxed{f(x)=\frac9{3x+1}-\frac{4x}{x^2+4}}.\]

(b) \[\begin{aligned}f(x)&=9(1-3x+9x^2-27x^3+\cdots)-x\left(1-\frac{x^2}{4}+\cdots\right)\\&=\boxed{9-28x+81x^2-\frac{971}{4}x^3+\cdots}.\end{aligned}\]

(c) The two expansions require \(|3x|<1\) and \(|x^2/4|<1\), respectively. Their common range is \(\boxed{|x|<1/3}\).

(a) The numerator and denominator have equal degree, so include a constant: \[f(x)=A+\frac B{x+3}+\frac{Cx+D}{x^2+2}.\] Multiplying by \((x+3)(x^2+2)\) gives \[x^3+2x-11=A(x+3)(x^2+2)+B(x^2+2)+(Cx+D)(x+3).\] Matching \(x^3\) gives \(A=1\). Substituting \(x=-3\) gives \(B=-4\). The remaining \(x^2\) and \(x\) coefficients give \(C=1,D=-3\). Thus \[f(x)=1-\frac4{x+3}+\frac{x-3}{x^2+2}.\]

(b) Expand each denominator: \[\frac1{x+3}=\frac13\left(1-\frac x3+\frac{x^2}{9}+\cdots\right),\quad\frac1{x^2+2}=\frac12\left(1-\frac{x^2}{2}+\cdots\right).\] Hence \[\begin{aligned}f(x)&=1-\frac43+\frac49x-\frac4{27}x^2+\frac x2-\frac32+\frac34x^2+\cdots\\&=-\frac{11}{6}+\frac{17}{18}x+\frac{65}{108}x^2+\cdots.\end{aligned}\] The common convergence range is \(|x|<\sqrt2\).

(a) Set \[f(x)=\frac A{3a+2x}+\frac B{2a-x}.\] Then \(3a-5x=A(2a-x)+B(3a+2x)\), so \(2A+3B=3\), \(-A+2B=-5\). Hence \[f(x)=\frac3{3a+2x}-\frac1{2a-x}.\]

(b) Rewrite and expand geometrically: \[\begin{aligned}f(x)&=\frac1a\left(1+\frac{2x}{3a}\right)^{-1}-\frac1{2a}\left(1-\frac{x}{2a}\right)^{-1}\\&=\frac1a\left(1-\frac{2x}{3a}+\frac{4x^2}{9a^2}\right)-\frac1{2a}\left(1+\frac{x}{2a}+\frac{x^2}{4a^2}\right)+\cdots\\&=\frac1{2a}-\frac{11x}{12a^2}+\frac{23x^2}{72a^3}+\cdots.\end{aligned}\]

(c) Both require \(|x|<3a/2\) and \(|x|<2a\), so the common interval is \[|x|<\frac{3a}2.\]

(a) Include a constant term: \[f(x)=A+\frac B{3x-2}+\frac C{x+6}.\] The leading coefficient gives \(A=4\). Subtracting the denominator times 4, \[-9x+46=B(x+6)+C(3x-2).\] At \(x=2/3\), \(B=6\); at \(x=-6\), \(C=-5\). Thus \[f(x)=4+\frac6{3x-2}-\frac5{x+6}.\]

(b) Expand as geometric series: \[\begin{aligned}f(x)&=4-3\left(1-\frac32x\right)^{-1}-\frac56\left(1+\frac x6\right)^{-1}\\&=4-3\left(1+\frac32x+\frac94x^2\right)-\frac56\left(1-\frac x6+\frac{x^2}{36}\right)+\cdots\\&=\frac16-\frac{157}{36}x-\frac{1463}{216}x^2+\cdots.\end{aligned}\] The common validity condition is \(|x|<2/3\).

(a) Write \(f(x)=A/(1+2x)+(Bx+C)/(2+x^2)\). Then \[5x^2+8x+5=A(2+x^2)+(Bx+C)(1+2x).\] Comparing coefficients gives \(A+2B=5,\ B+2C=8,\ 2A+C=5\), so \(A=1,\ B=2,\ C=3\). Hence \[f(x)=\frac1{1+2x}+\frac{2x+3}{2+x^2}.\]

(b) For \(|x|<1/2\), \[(1+2x)^{-1}=1-2x+4x^2-8x^3+\cdots,\] \[\frac{2x+3}{2+x^2}=\left(x+\frac32\right)\left(1-\frac{x^2}{2}+\cdots\right).\] The cubic contributions are −8 and −1/2, giving coefficient \(-17/2\).

(a) Setting \(f=A/(a-2x)+B/(3a+x)\) gives \(A=2a,B=a\). Hence \[f(x)=\frac{2a}{a-2x}+\frac a{3a+x}.\]

(b) \[f=2(1-2x/a)^{-1}+\frac13(1+x/(3a))^{-1}=\frac73+\frac{35}{9a}x+\frac{217}{27a^2}x^2+\cdots.\] (c) Both geometric expansions require \(|2x/a|<1\), \(|x/(3a)|<1\). Their intersection is \(-a/2<x<a/2\).