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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)\(\left(\frac{3}{4}+-\frac{7}{2}\right).\left(\frac{10}{11}+\frac{2}{22}\right)\)
\(=\left(\frac{3}{4}-\frac{14}{4}\right).\left(\frac{20}{22}+\frac{2}{22}\right)\)
\(=\left(-\frac{11}{4}\right).\left(\frac{22}{22}\right)\)
\(=-\frac{11}{4}\)
\(\frac{2^{19}.27^3+15.4^4.9^4}{6^9.2^{10}+12^{10}}=\frac{2^{19}.3^9+5.2^8.3^9}{2^9.3^9+2^{20}.3^{10}}=\frac{2^8.3^9\left(2^{11}-5\right)}{2^9.3^9\left(1+2^{11}.3\right)}=\frac{2043}{2.6145}=\frac{2043}{12290}\)
\(\frac{2^{19}\cdot27^3+15\cdot4^4\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
\(=\frac{2^{19}\cdot3^9+3\cdot5\cdot2^8\cdot3^8}{2^9\cdot3^9+2^{20}\cdot3^{10}}\)
\(=\frac{2^8\cdot3^9\left(2^{11}+5\right)}{2^9\cdot3^9\left(2^{11}\cdot3+1\right)}\)
\(=\frac{\left(2^{11}+5\right)}{2\left(2^{11}\cdot3+1\right)}\)
???
= 2100 - (299 + 298 + ... + 22 + 2)
Đặt A = 299 + 298 + ... + 22 + 2
2A = 2100 + 299 + ... + 23 + 22
2A - A = (2100 + 299 + ... + 23 + 22) - (299 + 298 + ... + 22 + 2)
A = 2100 - 2
Ta có:
2100 - 299 - 298 - ... - 22 - 2
= 2100 - (2100 - 2)
= 2100 - 2100 + 2
= 0 + 2
= 2
= 32 + 28 : 29
\(=9+\frac{1}{2}\)
\(=\frac{18}{2}+\frac{1}{2}=\frac{19}{2}\)