Tính M = \(\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}\)
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\(M=\sqrt{\dfrac{2^{30}-2^{20}}{2^{22}-2^{12}}}=\sqrt{\dfrac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}=\sqrt{2^8}=\sqrt{16^2}=16\)
\(M=\sqrt{\frac{8^{10}-4^{10}}{4^{11}-8^4}}\)
\(=\sqrt{\frac{2^{30}-2^{20}}{2^{22}-2^{12}}}\)
\(=\sqrt{\frac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}\)
\(=\sqrt{\frac{2^{20}}{2^{12}}}\)
\(=\sqrt{2^8}\)
\(=2^4\)
\(=16\)
=.= 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 \(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}=\sqrt{2^8}=\sqrt{\left(2^4\right)^2}=2^4=16\)
\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)
\(\dfrac{1}{\sqrt{11-2\sqrt{30}}}-\dfrac{3}{\sqrt{7-2\sqrt{10}}}+\dfrac{4}{\sqrt{8+4\sqrt{3}}}\)
\(=\dfrac{1}{\sqrt{11-2.\sqrt{6}.\sqrt{5}}}-\dfrac{3}{\sqrt{7-2.\sqrt{5}.\sqrt{2}}}+\dfrac{4}{\sqrt{2\left(4+2\sqrt{3}\right)}}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}}-\dfrac{3}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)2}}+\dfrac{4}{\sqrt{2\left(\sqrt{3}+1\right)^2}}\)
\(=\dfrac{1}{\sqrt{6}+\sqrt{5}}-\dfrac{3}{\sqrt{5}+\sqrt{2}}+\dfrac{2\sqrt{2}}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}-\dfrac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\dfrac{2\sqrt{2}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\sqrt{6}-\sqrt{5}+\sqrt{5}-\sqrt{2}+\sqrt{6}-\sqrt{2}=2\sqrt{6}-2\sqrt{2}\)
\(A=\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=\sqrt{\frac{2^{30}+2^{20}}{2^{22}+2^{12}}}=\sqrt{\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}}=\sqrt{\frac{2^{20}}{2^{12}}}=\sqrt{2^8}=\sqrt{\left(2^4\right)^2}\)\(=2^4=16.\)
#)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
\(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}\)
\(M=\sqrt{\dfrac{\left(2^3\right)^{10}-\left(2^2\right)^{10}}{\left(2^2\right)^{11}-\left(2^3\right)^4}}\)
\(M=\sqrt{\dfrac{2^{30}-2^{20}}{2^{22}-2^{12}}}\)
\(M=\sqrt{\dfrac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}\)
\(M=\sqrt{2^8}=16\)