Ta có :\(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)=\(\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)= \(\frac{2^{10}.\left(2^{10}+1\right)}{2^{12}.\left(2^{10}+1\right)}\)=\(\frac{2^{10}}{2^{12}}\)= 4.

phải là 1/4 chứ 

4^4

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

8^10+4^10/8^4+4^11

=(2^3)^10+(2^2)10/(2^3)^4+(2^2)^11

=2^30+2^20/2^12+2^22

=2^20(2^10+1)/2^12(1+2^10)

=2^8=256

vậy giá trị biểu thức =256

\(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)   

\(=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\)   

\(=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)   

\(=\frac{2^{30}+2^{20}}{2^{22}+2^{12}}\)   

\(=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}\)   

\(=\frac{2^{20}}{2^{12}}\)   

\(=2^8\)   

\(=256\)

\(A=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{12}.\left(2^{18}+2^8\right)}{2^{12}.\left(1+2^{10}\right)}=\frac{2^8.\left(2^{10}+1\right)}{1+2^{10}}=2^8\)

\(A=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)

\(A=\frac{2^{12}\cdot\left(2^{18}+2^8\right)}{2^{12}\cdot\left(1+2^{10}\right)}\)

\(A=\frac{2^8\cdot\left(2^{10}+1\right)}{2^{10}+1}\)

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

\(=2^8=256\)

Mình không phải CTV nhưng có thể giúp bạn  :)

Đừng dựa dẫm nhiều vào CTV nha bạn!

\(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)

\(=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)

\(=\frac{2^{20}×2^{10}+2^{20}}{2^{12}+2^{12}×2^{10}}\)

\(=\frac{2^{20}×\left(2^{10}+1\right)}{2^{12}×\left(1+2^{10}\right)}\)

\(=\frac{2^{20}}{2^{12}}=2^8\)

Cbht

#)Giải :

\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\frac{2^{20}}{2^{12}}=2^8=256\)

Ta có: \(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)=\(\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\)=\(\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)=\(\frac{2^{20}.2^{10}+2^{20}}{2^{12}.2^{10}+2^{12}}\)=\(\frac{2^{20}.\left(2^{10}+1\right)}{2^{12}.\left(2^{10}+1\right)}\)=\(\frac{2^{20}}{2^{12}}=2^8\)

bn ơi dễ như vậy mà không giải được sao

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