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

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

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

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

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

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

\(=2^4\)

\(=16\)

=.= hok tốt!!

Không dùng máy tính bỏ túi, tính \(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}\)

Ta có:

\(8^{10}-4^{10}=4^{10}\left(2^{10}-1\right)=4^6.4^4\left(2^{10}-1\right)=2^{12}.4^4\left(2^{10}-1\right)\)

\(4^{11}-8^4=4^4\left(4^7-2^4\right)=4^4\left(2^{14}-2^4\right)=4^4.2^4\left(2^{10}-1\right)\)

Do đó: \(\dfrac{8^{10}-4^{10}}{4^{11}-8^4}=\dfrac{2^{12}.4^4\left(2^{10}-1\right)}{2^4.4^4\left(2^{10}-1\right)}=\dfrac{2^{12}}{2^4}=2^{12-4}=2^8\)

Vậy \(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}=\sqrt{2^8}=\sqrt{\left(2^4\right)^2}=2^4=16\)

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

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

\(\dfrac{1}{\sqrt{11-2\sqrt{30}}}-\dfrac{3}{\sqrt{7-2\sqrt{10}}}+\dfrac{4}{\sqrt{8+4\sqrt{3}}}\)

\(=\dfrac{1}{\sqrt{11-2.\sqrt{6}.\sqrt{5}}}-\dfrac{3}{\sqrt{7-2.\sqrt{5}.\sqrt{2}}}+\dfrac{4}{\sqrt{2\left(4+2\sqrt{3}\right)}}\)

\(=\dfrac{1}{\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}}-\dfrac{3}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)2}}+\dfrac{4}{\sqrt{2\left(\sqrt{3}+1\right)^2}}\)

\(=\dfrac{1}{\sqrt{6}+\sqrt{5}}-\dfrac{3}{\sqrt{5}+\sqrt{2}}+\dfrac{2\sqrt{2}}{\sqrt{3}+1}\)

\(=\dfrac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}-\dfrac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\dfrac{2\sqrt{2}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\sqrt{6}-\sqrt{5}+\sqrt{5}-\sqrt{2}+\sqrt{6}-\sqrt{2}=2\sqrt{6}-2\sqrt{2}\)

\(A=\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=\sqrt{\frac{2^{30}+2^{20}}{2^{22}+2^{12}}}=\sqrt{\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}}=\sqrt{\frac{2^{20}}{2^{12}}}=\sqrt{2^8}=\sqrt{\left(2^4\right)^2}\)\(=2^4=16.\)

#)Giải :

\(A=\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=\sqrt{\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}}=\sqrt{\frac{2^{30}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}}=\sqrt{\frac{2^{30}}{2^{12}}}=\sqrt{2^8}=\sqrt{256}=16\)

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 ...

vô danh

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

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

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

\(M=\sqrt{\frac{2^{20}}{2^{12}}}\)

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

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

\(M=16\)

vậy \(M=16\)

P/S Đừng ai coppy bài mình nha