2) $$ \frac{2.3^{x}+3^{x-2}}{5 \cdot 3^{x+1}-7 \cdot 2^{x-1}} $$ 4) $$ \frac{9^{x}+3^{x}-2}{9^{x}-4} $$ 5) $$ 4^{x}+3 \cdot 2^{x+1} $$

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2) We start with
$$
\frac{2\cdot 3^{x}+3^{x-2}}{5\cdot 3^{x+1}-7\cdot 2^{x-1}}.
$$
In the numerator, factor out $$3^{x-2}$$:
$$
2\cdot 3^{x}+3^{x-2} = 3^{x-2}\Bigl(2\cdot 3^2+1\Bigr) = 3^{x-2}(18+1)=19\cdot 3^{x-2}.
$$
In the denominator, note that
$$
3^{x+1}=3^{x-2}\cdot3^3=27\cdot 3^{x-2},
$$
so we can write
$$
5\cdot 3^{x+1}-7\cdot 2^{x-1}=5\cdot 27\cdot 3^{x-2}-7\cdot 2^{x-1}=135\cdot 3^{x-2}-7\cdot2^{x-1}.
$$
Thus, the expression becomes
$$
\frac{19\cdot 3^{x-2}}{135\cdot 3^{x-2}-7\cdot 2^{x-1}}.
$$

4) We have
$$
\frac{9^{x}+3^{x}-2}{9^{x}-4}.
$$
Since $$9^x=(3^2)^x=3^{2x}$$, set $$t=3^x$$. Then the numerator becomes
$$
t^2+t-2,
$$
and the denominator becomes
$$
t^2-4.
$$
Factor both:
$$
t^2+t-2 = (t+2)(t-1) \quad \text{and} \quad t^2-4 = (t-2)(t+2).
$$
Cancel the common factor $$t+2$$ (which is always nonzero since $$3^x>0$$):
$$
\frac{(t+2)(t-1)}{(t-2)(t+2)}=\frac{t-1}{t-2}.
$$
Returning to $$t=3^x$$, we obtain
$$
\frac{3^x-1}{3^x-2}.
$$

5) We wish to simplify
$$
4^{x}+3\cdot 2^{x+1}.
$$
Write
$$
4^x = (2^2)^x = 2^{2x},
$$
and
$$
3\cdot 2^{x+1}= 3\cdot2\cdot2^{x}=6\cdot2^{x}.
$$
Then the expression becomes
$$
2^{2x}+6\cdot2^x.
$$
Factor out $$2^x$$:
$$
2^x(2^x+6).
$$
2) $$ \frac{19 \cdot 3^{x-2}}{135 \cdot 3^{x-2} - 7 \cdot 2^{x-1}} $$ 4) $$ \frac{3^x - 1}{3^x - 2} $$ 5) $$ 2^x(2^x + 6) $$

2) \( \frac{2.3^{x}+3^{x-2}}{5 \cdot 3^{x+1}-7 \cdot 2^{x-1}} \) 4) \( \frac{9^{x}+3^{x}-2}{9^{x}-4} \) 5) \( 4^{x}+3 \cdot 2^{x+1} \)

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