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a)|-10|:(-2):(-5)+(-3)2
=1+9
=10
b)1+(-2)+3+(-4)+5+(-6)+...+21+(-22)
=[1+(-2)]+[3+(-4)]+[5+(-6)]+...+[21+(-22]
=(-1)+(-1)+(-1)+...+(-1)
Mà từ 1 đến 22 có:(22-1):1+1:2=11(cặp)
Suy ra:1+(-2)+3+(-4)+5+(-6)+...+21+(-22)=(-11)
c)\(\frac{3}{4}.\frac{5}{9}+\frac{3}{4}.\frac{4}{9}\)
\(=\frac{3}{4}.\left(\frac{5}{9}+\frac{4}{9}\right)\)
\(=\frac{3}{4}\)
d)\(-\frac{4}{17}+\frac{5}{19}+-\frac{13}{17}+\frac{14}{19}+\frac{3}{115}\)
\(=\left[\left(-\frac{4}{17}\right)+\left(-\frac{13}{17}\right)\right]+\left(\frac{5}{19}+\frac{4}{19}\right)+\frac{3}{115}\)
\(=\left(-\frac{27}{17}\right)+1+\frac{3}{115}\)
\(=-\frac{1099}{1955}\)
e)\(\left(\frac{3}{4}+-\frac{7}{2}\right).\left(\frac{10}{11}+\frac{2}{22}\right)\)
\(=\left(\frac{3}{4}-\frac{14}{4}\right).\left(\frac{20}{22}+\frac{2}{22}\right)\)
\(=\left(-\frac{11}{4}\right).\left(\frac{22}{22}\right)\)
\(=-\frac{11}{4}\)
1) tự làm (thực hiện từ dưới lên)
2) B = \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{4}\right)^5.3}{\frac{\frac{1}{1024}.1}{3}-\left(\frac{1}{2}\right)^{11}}\)
= \(\frac{\left(\frac{1}{2}\right)^{10}.5-\left(\frac{1}{2}\right)^{10}.3}{\left(\frac{1}{2}\right)^{10}.\frac{1}{3}-\left(\frac{1}{2}\right)^{10}.\frac{1}{2}}\)
= \(\frac{\left(\frac{1}{2}\right)^{10}.\left(5-3\right)}{\left(\frac{1}{2}\right)^{10}.\left(\frac{1}{3}-\frac{1}{2}\right)}\)
= \(\frac{2}{-\frac{1}{6}}\)= 2 . (-6) = -12
1) \(5+\frac{1}{1+\frac{1}{1+\frac{2}{1+\frac{3}{4}}}}=5+\frac{15}{7}=\frac{5}{1}+\frac{15}{7}=\frac{50}{7}\)
đặt \(A=\frac{7}{10}+\frac{7}{10^2}+\frac{7}{10^3}+\frac{7}{10^4}\)
\(A=7.\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)
Lại đặt \(B=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\)
\(10B=1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\)
\(10B-B=\left(1+\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}\right)-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}\right)\)
\(9B=1-\frac{1}{10^4}\)
\(\Rightarrow B=\frac{1-\frac{1}{10^4}}{9}\)
\(\Rightarrow A=7.\frac{1-\frac{1}{10^4}}{9}=\frac{7.\left(1-\frac{1}{10^4}\right)}{9}\)
\(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3 \left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\)
=3[1/5+1/7-1/11]
----------------------- =3/4
4[1/5+1/7-1/11]
Bài giải
\(d,\text{ }\frac{72^3\cdot54^2}{108^4}=\frac{3^6\cdot2^9\cdot3^6\cdot2^2}{3^{12}\cdot2^8}=2^3=9\)
\(e,\text{ }\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\frac{3^{10}\left(11+5\right)}{3^9\cdot2^4}=\frac{3\cdot16}{2^4}=\frac{3\cdot2^4}{2^4}=3\)