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\(\frac{85-17+34}{51-102}\)
\(=\frac{17\left(5-1+2\right)}{51\left(1-2\right)}\)
\(=\frac{17.6}{51.\left(-1\right)}\)
\(=\frac{1.6}{3.\left(-1\right)}\)
\(=\frac{1.2}{1.\left(-1\right)}\)
\(=\frac{2}{-1}\)
\(=-2\)
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) \(\frac{85-17+34}{51-102}=-2\)
a. 3.5.11.13 3..5.11.13 13 13
33.35.37 = 3.11.5.7.37 = 7.37 = 259
Rút gọn phân số :
a) \(\dfrac{85-17+34}{51-102}\)
\(=\dfrac{102}{-51}=-2\)
b) \(\dfrac{-13.6+12.5}{6\left(-7\right)-\left(-4\right).6}\)
\(=\dfrac{-78+60}{6\left[\left(-7\right)-\left(-4\right)\right]}\)
\(=\dfrac{-18}{-18}=1\)
c) \(\dfrac{2^4.5^2.7}{2^3.5.7^2.11}\)
\(=\dfrac{2^4}{2^3}.\dfrac{5^2}{5}.\dfrac{7}{7^2}.11\)
\(=2.5.\dfrac{1}{7}.11\)
\(=15\dfrac{5}{7}=\dfrac{110}{7}\)
\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(=\frac{17.2}{7.13}+\frac{17.3}{13.22}+\frac{17.5}{22.37}+\frac{17.4}{37.49}\)
\(=17\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+...+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)\(=\frac{17}{3}.\frac{6}{49}=\frac{17.2}{49}=\frac{34}{49}\)
Có : \(\frac{17}{24}=\frac{34}{48}\)
Vì 48 < 49 => \(\frac{34}{48}>\frac{34}{49}\). Hay \(\frac{17}{24}>A\)
\(\frac{85-17+34}{51-102}=\frac{102}{-51}=-2\)
Ta thấy:
\(\frac{85-17+34}{51-102}=\frac{17\cdot5-17+17\cdot2}{17\cdot3-17\cdot6}=\frac{17\cdot\left(5-1+2\right)}{17\cdot\left(3-6\right)}=\frac{17\cdot\left(4+2\right)}{17\cdot\left[3+\left(-6\right)\right]}=\frac{17\cdot6}{17\cdot-3}=-2\)