\(y=\frac{16.29-32}{16.37+32}=\frac{9}{13}\)

vậy y = \(\frac{9}{13}\)

t i ck nha!! 6546576876876976

= 29/37

\(\frac{16.29-32}{16.37+32}\)

\(=\frac{16.29-16.2}{16.37+16.2}\)

\(=\frac{16\left(29-2\right)}{16\left(37+2\right)}\)

\(=\frac{16.27}{16.39}\)

\(=\frac{27}{39}=\frac{9}{13}\)

a) \(\frac{3^2.2^4}{8.3^2}=\frac{3^2.2^4}{2^3.3^2}=2\)

b)\(\frac{84.45}{49.54}=\frac{2^2.7.3^3.5}{7^2.2.3^3}=\frac{10}{7}\)

c) \(\frac{36.10.21}{2^3.63.35}=\frac{2^3.3^3.5.7}{2^3.3^2.5.7^2}=\frac{3}{7}\)

d) \(\frac{16.29-32}{16.37+32}=\frac{16.29-16.2}{16.37+2.16}=\frac{16.\left(29-2\right)}{16.\left(37+2\right)}=\frac{27}{39}=\frac{9}{13}\)

3².2⁴

= 3².2².2²

= 6².2²

= 12²

-4165/5

\(\dfrac{-270}{450}=\dfrac{-9}{15}\)
\(\dfrac{11}{-143}=\dfrac{1}{-13}\)
\(\dfrac{32}{12}=\dfrac{8}{3}\)
\(\dfrac{-26}{-156}=\dfrac{1}{6}\)

a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)

\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)

\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)

\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)

\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)

\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)

\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)

\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)

\(\dfrac{-351}{702}=\dfrac{-1}{2}\)

\(-\dfrac{18181818}{19191919}=\dfrac{-18}{19}\)

\(\dfrac{-583}{352}=-\dfrac{53}{32}\)

-2/3

7/13

3/5

TK
-2/3

7/13

3/5