f) \(\left(1:\frac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\frac{1}{49}\)

\(=\left(1\cdot7\right)^2:\left(2^6:2^5\right)\cdot\frac{1}{49}=7^2\cdot\frac{1}{2}\cdot\frac{1}{49}=49\cdot\frac{1}{49}\cdot\frac{1}{2}=\frac{1}{2}\)

g) \(\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{\left(2^2\right)^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot\left(3^2\right)^3+\left(2^3\right)^4\cdot3^5}\)

\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\left(3^5-3^6\right)}{2^{12}\left(3^6+3^5\right)}=\frac{2^{12}\left[3^5\left(1-3\right)\right]}{2^{12}\left[3^5\left(3+1\right)\right]}=\frac{2^{12}\cdot3^5\cdot\left(-2\right)}{2^{12}\cdot3^5\cdot4}=\frac{-2}{4}=-\frac{1}{2}\)

                                                               Bài giải

\(f,\text{ }\left(1\text{ : }\frac{1}{7}\right)^2\left[\left(2^2\right)^3\text{ : }2^5\right]\cdot\frac{1}{49}\)

\(=7^2\left(2^6\text{ : }2^5\right)\cdot\frac{1}{7^2}\)

\(=2\)

\(g,\text{ }\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^5\cdot\left(1-3\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}=-\frac{2}{4}=-\frac{1}{2}\)

mk ko viết lại đề

\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)

\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)

\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)

Vậy A= \(\frac{1}{2}\)

\(\frac{2^{12}\cdot3^5-\left(2^2\right)^6.3^5.3}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^5}\)

=\(\frac{2^{12}\cdot3^5-2^{12}.3^5.3}{2^{12}.3^5+2^{12}.3^5}\)

=3

Ta có:

\(A=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^{19}.3^{19}.2^9-7.2^{29}.3^{18}}\)

\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}=\frac{2^{29}.3^{18}\left(2.5-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}\)

\(=\frac{2^{29}.3^{18}}{2^{28}.3^{18}}=2\)

k nha

\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)

\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)