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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) \(\frac{-28}{72}=\frac{-7}{18}\)
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)\(\frac{-315}{540}\)=\(\frac{7}{12}\)
b)\(\frac{25.13}{26.35}\)=\(\frac{5}{14}\)
c)\(\frac{6.9-2.17}{63.3-119}\)=\(\frac{4}{7}\)
d)\(\frac{2929-101}{2.1919+404}\)=\(\frac{2}{3}\)
a) \(\dfrac{2\cdot7\cdot13}{26\cdot35}=\dfrac{2\cdot7\cdot13}{2\cdot13\cdot5\cdot7}=\dfrac{1}{5}\)
b) \(\dfrac{31\cdot7-13}{4-35}=\dfrac{217-13}{-31}=\dfrac{-204}{31}\)
c) \(\dfrac{32\cdot7-32\cdot4}{38-35}=\dfrac{32\cdot\left(7-4\right)}{3}=\dfrac{32\cdot3}{3}=32\)
d) \(\dfrac{1717-101}{2828+202}=\dfrac{1616}{3030}=\dfrac{8}{15}\)
e) \(\dfrac{3\cdot6+2\cdot9\cdot5-18\cdot\left(-4\right)}{7\cdot\left(-7\right)+12\cdot\left(-7\right)+7}\)
\(=\dfrac{18+18\cdot5+18\cdot4}{7\cdot\left(-7\right)-12\cdot7+7}\)
\(=\dfrac{18\cdot\left(1+5+4\right)}{7\cdot\left(-7-12+1\right)}\)
\(=\dfrac{18\cdot10}{7\cdot\left(-18\right)}\)
\(=\dfrac{-10}{7}\)
\(-\frac{3}{6}=\frac{x}{-2}=-\frac{18}{y}=-\frac{z}{24}\)
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: \(\dfrac{-315}{540}=\dfrac{-315:45}{540:45}=\dfrac{-7}{12}\)
b: \(\dfrac{25\cdot13}{26\cdot35}=\dfrac{25}{35}\cdot\dfrac{1}{2}=\dfrac{5}{7}\cdot\dfrac{1}{2}=\dfrac{5}{14}\)
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) \(\frac{1717-101}{2828+404}=\frac{101.17-101}{404.7+404}=\frac{101.\left(17-1\right)}{404.\left(7+1\right)}=\frac{101.16}{404.8}=\frac{101.2.8}{101.4.8}=\frac{1}{2}\)