Thực hiện phép tính
\(\frac{2^4×2^6}{\left(2^5\right)^2}\)-\(\frac{2^5×15^3}{6^3×10^2}\)
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Ta có: \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)....\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}...\frac{779}{780}\)
\(=\frac{4}{6}.\frac{10}{12}....\frac{1558}{1560}\)
\(=\frac{1.4.2.5....38.41}{2.3.3.4....39.40}=\frac{\left(1.2.3..38\right)\left(4.5...41\right)}{\left(2.3.4...39\right)\left(3...40\right)}=\frac{41}{39.3}=\frac{41}{117}\)
\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}........\frac{779}{780}\)
\(=\frac{4}{6}.\frac{10}{12}\frac{18}{20}.\frac{28}{30}.........\frac{1558}{1560}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...............\frac{38.41}{39.40}\)
\(=\frac{\left(1.2.3.4......38\right)\left(4.5.6.7..........41\right)}{\left(2.3.4.5.........39\right)\left(3.4.5.6.........40\right)}\)
\(=\frac{1.41}{39.3}\)
\(=\frac{41}{117}\)
Vậy \(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)........\left(1-\frac{1}{780}\right)=\frac{41}{117}\)
\(a,\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)
\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}=\frac{-7}{25}-\frac{18}{25}=-1\)
\(b,\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.0=0\)
c)\(5\frac{2}{5}.4\frac{2}{7}+5\frac{5}{7}.5\frac{2}{5}=\frac{27}{5}.\frac{30}{7}+\frac{40}{7}.\frac{27}{5}=\frac{27}{5}.\left(\frac{30}{7}+\frac{40}{7}\right)\)
\(=\frac{27}{5}.10=27.2=54\)
\(d,75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2=\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{12}{5}-\frac{1}{4}\)
\(=\left(\frac{3}{4}-\frac{1}{4}\right)-\frac{3}{2}+\frac{6}{5}=\frac{1}{2}-\frac{3}{2}+\frac{6}{5}=-1+\frac{6}{5}=\frac{-5}{5}+\frac{6}{5}=\frac{1}{5}\)
\(A=\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{\frac{9-2.3\sqrt{3}+3}{2}}=\frac{\sqrt{2}\left(3+\sqrt{3}\right)}{\sqrt{2}}.\sqrt{\left(3-\sqrt{3}\right)^2}=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
chia ra làm các nhóm (13-12+11)+(10-9+8)-(7-6+5) -(4+3+2-1)
hay ta thấy 3 nhóm đầu có tổng =số trừ trong các nhóm đó tức là 12+9-6+(4+3+2-1)
Hay= 23
\(\frac{2^{4+6}}{2^{5.2}}-\frac{2^5.3^3.5^3}{2^3.3^3.2^2.5^2}=\frac{2^{10}}{2^{10}}-\frac{2^5.3^3.5^3}{2^5.3^3.5^2}=1-5=-4\)
bn hoc gioi thiet