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9 tháng 8 2018
\(M=\sqrt{\frac{8^{10}-4^{10}}{4^{11}-8^4}}\)
\(=\sqrt{\frac{2^{30}-2^{20}}{2^{22}-2^{12}}}\)
\(=\sqrt{\frac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}\)
\(=\sqrt{\frac{2^{20}}{2^{12}}}\)
\(=\sqrt{2^8}\)
\(=2^4\)
\(=16\)
=.= hok tốt!!
TH
1
20 tháng 7 2017
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 \(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}=\sqrt{2^8}=\sqrt{\left(2^4\right)^2}=2^4=16\)
TN
1
17 tháng 11 2021
\(M=\sqrt{\dfrac{2^{30}-2^{20}}{2^{22}-2^{12}}}=\sqrt{\dfrac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}=\sqrt{2^8}=\sqrt{16^2}=16\)
M
20 tháng 11 2017
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
5 tháng 11 2017
\(\dfrac{1}{\sqrt{11-2\sqrt{30}}}-\dfrac{3}{\sqrt{7-2\sqrt{10}}}+\dfrac{4}{\sqrt{8+4\sqrt{3}}}\)
\(=\dfrac{1}{\sqrt{11-2.\sqrt{6}.\sqrt{5}}}-\dfrac{3}{\sqrt{7-2.\sqrt{5}.\sqrt{2}}}+\dfrac{4}{\sqrt{2\left(4+2\sqrt{3}\right)}}\)
\(=\dfrac{1}{\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}}-\dfrac{3}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)2}}+\dfrac{4}{\sqrt{2\left(\sqrt{3}+1\right)^2}}\)
\(=\dfrac{1}{\sqrt{6}+\sqrt{5}}-\dfrac{3}{\sqrt{5}+\sqrt{2}}+\dfrac{2\sqrt{2}}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{6}-\sqrt{5}}{\left(\sqrt{6}+\sqrt{5}\right)\left(\sqrt{6}-\sqrt{5}\right)}-\dfrac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\dfrac{2\sqrt{2}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\sqrt{6}-\sqrt{5}+\sqrt{5}-\sqrt{2}+\sqrt{6}-\sqrt{2}=2\sqrt{6}-2\sqrt{2}\)
9 tháng 12 2022
\(C=\dfrac{\sqrt{10}-\sqrt{5}+2\sqrt{2}+\sqrt{5}-\sqrt{10}-1}{2\sqrt{2}+2+2\sqrt{2}-1+2\sqrt{2}+2}\)
\(=\dfrac{2\sqrt{2}-1}{6\sqrt{2}+3}=\dfrac{9-4\sqrt{2}}{21}\)
\(B=\dfrac{40}{6+2\sqrt{5}+\sqrt{4\sqrt{5}+4}}\)
\(=\dfrac{40}{\left(\sqrt{5}+1\right)^2+2\sqrt{\sqrt{5}+1}}\)
\(=\dfrac{40}{\sqrt{\sqrt{5}+1}\left(\sqrt{\sqrt{5}+1}+2\right)}\)
\(=\dfrac{40\sqrt{\sqrt{5}-1}}{2\left(\sqrt{\sqrt{5}+1}+2\right)}\)
\(=\dfrac{20\left(\sqrt{\sqrt{5}-1}\right)\left(\sqrt{\sqrt{5}+1}-2\right)}{\sqrt{5}+1-4}\)
\(=\dfrac{20\left(\sqrt{\sqrt{5}-1}\right)\left(\sqrt{\sqrt{5}+1}-2\right)}{-3+\sqrt{5}}\)
\(=-5\left(3+\sqrt{5}\right)\left(\sqrt{\sqrt{5}-1}\right)\left(\sqrt{\sqrt{5}+1}-2\right)\)
15 tháng 7 2018
a, ta có
\(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}< 3+4< 7\) (1)
lại có \(\sqrt{65}-1>\sqrt{64}-1>8-1>7\) (2)
từ (1) và(2) =>\(\sqrt{8}+\sqrt{15}< \sqrt{65}-1\)
bài 2
\(M=\sqrt{\frac{\left(2^3\right)^{10}-\left(2^2\right)^{10}}{\left(2^2\right)^{11}-\left(2^3\right)^4}}=\sqrt{\frac{2^{30}-2^{20}}{2^{22}-2^{12}}}=\sqrt{\frac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}=\sqrt{\frac{2^{20}}{2^{12}}}=\sqrt{2^8}=2^4\)
7 tháng 8 2018
1/ Tính: \(A=\dfrac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{10}+1\right)^2}}{2\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}+\sqrt{\left(2\sqrt{2}+2\right)^2}}=\dfrac{\sqrt{10}-\sqrt{5}+2\sqrt{2}+\sqrt{5}-\sqrt{10}-1}{2\sqrt{2}+2+2\sqrt{2}-1+2\sqrt{2}+2}=\dfrac{2\sqrt{2}-1}{6\sqrt{2}-3}=\dfrac{2\sqrt{2}-1}{3\left(2\sqrt{2}-1\right)}=\dfrac{1}{3}\)
7 tháng 8 2018
\(B=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2}+\sqrt{3}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2}-\sqrt{3}}=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2}-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2}-\sqrt{3}\right)}=\dfrac{2\sqrt{2}-2\sqrt{2}-2\sqrt{3}+\sqrt{6}-\sqrt{6}-3+2\sqrt{2}+2\sqrt{2}+2\sqrt{3}-\sqrt{6}-\sqrt{6}-3}{2-\left(\sqrt{2}+\sqrt{3}\right)^2}=\dfrac{4\sqrt{2}-2\sqrt{6}-6}{2-2-3-2\sqrt{6}}=\dfrac{2\left(2\sqrt{2}-\sqrt{6}-3\right)}{-3-2\sqrt{6}}\)
\(M=\sqrt{\dfrac{8^{10}-4^{10}}{4^{11}-8^4}}\)
\(M=\sqrt{\dfrac{\left(2^3\right)^{10}-\left(2^2\right)^{10}}{\left(2^2\right)^{11}-\left(2^3\right)^4}}\)
\(M=\sqrt{\dfrac{2^{30}-2^{20}}{2^{22}-2^{12}}}\)
\(M=\sqrt{\dfrac{2^{20}\left(2^{10}-1\right)}{2^{12}\left(2^{10}-1\right)}}\)
\(M=\sqrt{2^8}=16\)