\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+3^{11}.2^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2.6}{3.7}=\frac{12}{21}=\frac{4}{7}\)

\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{4}{7}\)

ngọc diệp làm thế thì ai hiểu

\(y=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)

\(y=\frac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(y=\frac{2^9.3^9\left(2^3.3+120\right)}{2^{11}.3^{11}\left(2.3+1\right)}\)

\(y=\frac{6^9\left(2^3.3+120\right)}{6^{11}.7}\)

\(y=\frac{2^3.3+120}{6^2.7}\)

\(y=\frac{144}{252}\)

\(y=\frac{4}{7}\)

Lời giải:

Gọi biểu thức là $A$.

\(A=\frac{(2^4)^3.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\\ =\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}(2.3+1)}\\ =\frac{2^{12}.3^{10}(1+5)}{7.2^{11}.3^{11}}=\frac{2^{12}.3^{10}.2.3}{7.2^{11}.3^{11}}\\ =\frac{2^{13}.3^{11}}{7.2^{11}.3^{11}}=\frac{2^2}{7}=\frac{4}{7}\)

\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+3^{11}.2^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2.6}{3.7}=\frac{12}{21}=\frac{4}{7}\)

=\(\frac{2^{13}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

=\(\frac{2^{12}\cdot3^{10}\cdot\left(1+2\cdot5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3+1\right)}\)

=\(\frac{2\cdot11}{3\cdot7}\)

duyệt nha các bn

=\(\frac{22}{21}\)

\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)

=\(\dfrac{\left(4^2\right)^3.3^{10}+120.1}{4^6.3^{12}+6^{11-9}}\)

=\(\dfrac{4^{2.3}.1+120}{4^6.3^{12-10}+6^2}\)

=\(\dfrac{4^6+120}{4^6.3^2+6^2}\)

=\(\dfrac{4096+120}{4096.9+36}\)

=\(\dfrac{4216}{36864+36}\)

=\(\dfrac{4216}{36900}=\dfrac{2063}{18450}\)