Bài tập: tính giá trị biểu thức
\(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)
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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.
\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=2^8\)
\(A=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{12}.\left(2^{18}+2^8\right)}{2^{12}.\left(1+2^{10}\right)}=\frac{2^8.\left(2^{10}+1\right)}{1+2^{10}}=2^8\)
\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}\)
\(=2^8=256\)
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\(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)
\(=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)
\(=\frac{2^{20}×2^{10}+2^{20}}{2^{12}+2^{12}×2^{10}}\)
\(=\frac{2^{20}×\left(2^{10}+1\right)}{2^{12}×\left(1+2^{10}\right)}\)
\(=\frac{2^{20}}{2^{12}}=2^8\)
Cbht
Ta có: \(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)=\(\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\)=\(\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)=\(\frac{2^{20}.2^{10}+2^{20}}{2^{12}.2^{10}+2^{12}}\)=\(\frac{2^{20}.\left(2^{10}+1\right)}{2^{12}.\left(2^{10}+1\right)}\)=\(\frac{2^{20}}{2^{12}}=2^8\)
Ta có:
\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=2^8=256\)
\(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)
\(=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\)
\(=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\)
\(=\frac{2^{30}+2^{20}}{2^{22}+2^{12}}\)
\(=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}\)
\(=\frac{2^{20}}{2^{12}}\)
\(=2^8\)
\(=256\)