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a. 3.5.11.13 3..5.11.13 13 13
33.35.37 = 3.11.5.7.37 = 7.37 = 259
a) \(\frac{2.7.13}{26.35}=\frac{2.7.13}{2.13.5.7}=\frac{1}{5}\)
b) \(\frac{23.5-23}{4-27}=\frac{23.\left(5-1\right)}{-23}=\frac{23.4}{-23}=\frac{23.\left(-4\right)}{23}=-4\)
c) \(\frac{2130-15}{3550-25}=\frac{142.15-15}{142.25-25}=\frac{15.\left(142-1\right)}{25.\left(142-1\right)}=\frac{15.141}{25.141}=\frac{15}{25}=\frac{3.5}{5.5}=\frac{3}{5}\)
d) \(\frac{1717-101}{2828+404}=\frac{17.101-101}{28.101+101.4}=\frac{101.\left(17-1\right)}{101.\left(28+4\right)}=\frac{101.16}{101.32}=\frac{16}{32}=\frac{4.4}{2.4.4}=\frac{1}{2}\)
e) \(\frac{3.5.11.13}{33.35.27}=\frac{3.5.11.13}{3.11.5.7.3^3}=\frac{13}{7.3^3}=\frac{13}{189}\)
f) \(\frac{85-17+34}{51-102}=\frac{5.17-17+2.17}{3.17-6.17}=\frac{17.\left(5-1+2\right)}{17.\left(3-6\right)}=\frac{17.6}{17.\left(-3\right)}=\frac{6}{-3}=-2\)
a) \(\frac{2.7.13}{26.35}=\frac{2.7.13}{2.13.7.5}=\frac{1}{5}\)
b) \(\frac{3.5.11.13}{33.35.37}=\frac{3.5.11.13}{3.11.7.5.37}=\frac{13}{259}\)
c) \(\frac{23.5-23}{4-27}=\frac{23.\left(5-1\right)}{-23}=\frac{4}{-1}=-4\)
d) \(\frac{1717-101}{2828+404}=\frac{101.17-101}{404.7+404}=\frac{101.\left(17-1\right)}{404.\left(7+1\right)}=\frac{101.16}{404.8}=\frac{101.2.8}{101.4.8}=\frac{1}{2}\)
Ta thấy:
\(\frac{85-17+34}{51-102}=\frac{17\cdot5-17+17\cdot2}{17\cdot3-17\cdot6}=\frac{17\cdot\left(5-1+2\right)}{17\cdot\left(3-6\right)}=\frac{17\cdot\left(4+2\right)}{17\cdot\left[3+\left(-6\right)\right]}=\frac{17\cdot6}{17\cdot-3}=-2\)
\(-\frac{3}{6}=\frac{x}{-2}=-\frac{18}{y}=-\frac{z}{24}\)
Ta có :+) \(-\frac{3}{6}=\frac{x}{-2}\)
\(\Rightarrow x=\frac{\left(-3\right)\left(-2\right)}{6}\)
\(\Rightarrow x=1\)
+)\(-\frac{3}{6}=-\frac{18}{y}\)
\(\Rightarrow y=\frac{6.\left(-18\right)}{-3}\)
\(\Rightarrow y=36\)
+)\(-\frac{3}{6}=-\frac{z}{24}\)
\(\Rightarrow-z=\frac{\left(-3\right)24}{6}\)
\(\Rightarrow-z=-12\)
\(\Rightarrow z=12\)
Vậy........................
\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=\frac{17}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)
\(=13\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)
\(=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)\)
\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)
\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}{\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)
a) \(\frac{-28}{72}=\frac{-7}{18}\)
b) \(\frac{-75}{-105}=\frac{75}{105}=\frac{5}{7}\)
c) \(\frac{3.5.11.13}{33.35.37}=\frac{3.5.11.13}{3.11.5.7.37}=\frac{13}{7.37}=\frac{13}{259}\)
d) \(\frac{85-17+34}{51-102}=\frac{102}{-51}=-2\)
a) \(\frac{-28}{72}=\frac{\left(-28\right):4}{72:4}=\frac{-7}{18}\)
b) \(\frac{-75}{-105}=\frac{\left(-75\right):\left(-15\right)}{\left(-105\right):\left(-15\right)}=\frac{5}{7}\)
c) \(\frac{3.5.11.13}{33.35.37}=\frac{3.5.11.13}{11.3.5.7.37}=\frac{13}{7.37}=\frac{13}{259}\)
d) \(\frac{85-17+34}{51-102}=-2\)