Simplify the following (leave your answers with positive exponents): \( \begin{array}{ll}\text { a) } \frac{3^{x+1}-2.3^{x}}{3^{x} \cdot 2} & \text { b) } \frac{2^{2 x+3}-5.2^{2 x+1}}{4^{x+2}} \\ \text { c) } \frac{35^{a}-3.5^{a}}{2^{2 a} 7^{a}-3.2^{2 a}} & \text { d) } \frac{3^{a+1} \cdot 4^{a}+5.3^{a+1}}{4^{2 a}-25} \\ \text { e) } \frac{7^{a} \cdot 49-7^{a+2} \cdot 2^{-1}}{2^{-3} \cdot 7^{a}} & \text { f) } \frac{2^{3 a-1}+\frac{3}{2}}{2^{4 a-1}+3.2^{a-1}} \\ \text { g) } \frac{-2^{3}-2^{5}}{5^{a+1} \cdot 2^{2}} & \text { h) } \frac{\frac{3}{4^{2}+4^{a-2}}}{3.4^{3}+4^{a+3}}\end{array} \)

a) We have \[ \frac{3^{x+1}-2\cdot3^{x}}{3^x\cdot2}. \] Factor \(3^x\) in the numerator: \[ 3^x\Bigl(3-2\Bigr)=3^x\cdot1. \] Thus, the expression becomes \[ \frac{3^x}{2\cdot3^x}=\frac{1}{2}. \] --- b) We are to simplify \[ \frac{2^{2x+3}-5\cdot2^{2x+1}}{4^{x+2}}. \] Express the denominator in base 2: \[ 4^{x+2}=(2^2)^{x+2}=2^{2x+4}. \] Factor \(2^{2x+1}\) from the numerator: \[ 2^{2x+3}-5\cdot2^{2x+1}=2^{2x+1}\Bigl(2^2-5\Bigr)=2^{2x+1}(4-5)=-2^{2x+1}. \] Hence, the fraction becomes \[ \frac{-2^{2x+1}}{2^{2x+4}}=-2^{\, (2x+1)-(2x+4)}=-2^{-3}=-\frac{1}{2^3}=-\frac{1}{8}. \] --- c) We wish to simplify \[ \frac{35^a-3\cdot5^a}{2^{2a}7^a-3\cdot2^{2a}}. \] Write \(35^a\) as \((5\cdot7)^a=5^a7^a\). Then \[ 35^a-3\cdot5^a=5^a7^a-3\cdot5^a=5^a\Bigl(7^a-3\Bigr). \] Similarly, factor \(2^{2a}\) in the denominator: \[ 2^{2a}7^a-3\cdot2^{2a}=2^{2a}\Bigl(7^a-3\Bigr). \] Cancel the common factor \(7^a-3\) (assuming \(7^a-3\neq0\)): \[ \frac{5^a}{2^{2a}}=\frac{5^a}{(2^2)^a}=\left(\frac{5}{4}\right)^a. \] --- d) We simplify \[ \frac{3^{a+1}\cdot 4^a+5\cdot3^{a+1}}{4^{2a}-25}. \] Factor \(3^{a+1}\) in the numerator: \[ 3^{a+1}\Bigl(4^a+5\Bigr). \] Recognize the denominator as a difference of squares: \[ 4^{2a}-25=\left(4^a\right)^2-5^2=(4^a-5)(4^a+5). \] Cancel the common factor \(4^a+5\): \[ \frac{3^{a+1}}{4^a-5}. \] --- e) We simplify \[ \frac{7^a\cdot49-7^{a+2}\cdot2^{-1}}{2^{-3}\cdot7^a}. \] Notice that \[ 7^a\cdot49=7^a\cdot7^2=7^{a+2}. \] Thus the numerator becomes \[ 7^{a+2}-7^{a+2}\cdot2^{-1}=7^{a+2}\Bigl(1-\frac{1}{2}\Bigr)=7^{a+2}\cdot\frac{1}{2}. \] Now, the denominator is \[ 2^{-3}\cdot7^a=\frac{7^a}{8}. \] Thus, \[ \frac{7^{a+2}\cdot\frac{1}{2}}{\frac{7^a}{8}}=\frac{7^{a+2}}{2}\cdot\frac{8}{7^a}=\frac{8\cdot7^{a+2}}{2\cdot7^a}=\frac{8\cdot7^2}{2}=\frac{8\cdot49}{2}=196. \] --- f) We simplify \[ \frac{2^{3a-1}+\frac{3}{2}}{2^{4a-1}+3\cdot2^{a-1}}. \] Write \(\frac{3}{2}\) as \(3\cdot2^{-1}\). Then the numerator is \[ 2^{3a-1}+3\cdot2^{-1}=2^{-1}\Bigl(2^{3a}+3\Bigr). \] Similarly, factor \(2^{a-1}\) from the denominator: \[ 2^{4a-1}+3\cdot2^{a-1}=2^{a-1}\Bigl(2^{3a}+3\Bigr). \] Cancel the common factor \(2^{3a}+3\) (assuming it is nonzero): \[ \frac{2^{-1}}{2^{a-1}}=2^{-1-(a-1)}=2^{-a}. \] --- g) We simplify \[ \frac{-2^3-2^5}{5^{a+1}\cdot2^2}. \] Evaluate the numerator: \[ -2^3-2^5=-(2^3+2^5)=-(8+32)=-40. \] The denominator is \[ 5^{a+