\(M=\sqrt{\dfrac{2^{30}-2^{20}}{2^{22}-2^{12}}}=\sqrt{\dfrac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}=\sqrt{2^8}=\sqrt{16^2}=16\)

\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)

=\(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=\dfrac{11}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)

Giải:

\(M=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}\)

\(\Leftrightarrow M=\dfrac{2^{10}.4^{10}+4^{10}}{2^4.4^4+4^{11}}\)

\(\Leftrightarrow M=\dfrac{4^{10}\left(2^{10}+1\right)}{4^4.2^4\left(2^{10}+1\right)}\)

\(\Leftrightarrow M=\dfrac{4^6}{2^4}\)

\(\Leftrightarrow M=\dfrac{2^{12}}{2^4}\)

\(\Leftrightarrow M=2^8=256\)

Vậy ...

\(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}\)

\(M=\sqrt{\dfrac{\left(2^3\right)^{10}-\left(2^2\right)^{10}}{\left(2^2\right)^{11}-\left(2^3\right)^4}}\)

\(M=\sqrt{\dfrac{2^{30}-2^{20}}{2^{22}-2^{12}}}\)

\(M=\sqrt{\dfrac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}\)

\(M=\sqrt{2^8}=16\)

a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)

b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)

c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)

\(A=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=2^8\)

Vậy...

M=256

N=15^15/3^15

Thông cảm vì mình ko giải ra chi tiết vì nó lâuuuu

N = \(\dfrac{3^{30}.5^{30}}{3^{30}.5^{15}}=\dfrac{5^{30}}{5^{15}}=5^{15}\)

\(M=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=2^8=256\\ N=\dfrac{\left(3\cdot5\right)^{30}}{\left(3^2\cdot5\right)^{15}}=\dfrac{3^{30}\cdot5^{30}}{3^{30}\cdot5^{15}}=5^{15}=...\)

\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=2^8=256\)

chỉ cách tính hay là có cần tính kết quả luôn k bn

Mình ns kết quả nha

Bằng 256

Ủng hộ mk nhaaa