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a ) \(\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{-12}{19}\)
\(=\frac{7}{19}.\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{-12}{19}\)
\(=\frac{7}{19}.\frac{11}{11}+\frac{-12}{19}\)
\(=\frac{7}{19}.1+\frac{-12}{19}\)
\(=\frac{7}{19}+\frac{-12}{19}\)
\(=\frac{7+\left(-12\right)}{19}\)
\(=-\frac{5}{19}\)
b ) \(\frac{-7}{25}.\frac{39}{-14}.\frac{50}{78}=\frac{-7.39.50}{25.-14.78}=\frac{-1.1.2}{1.-2.2}=\frac{-2}{-4}=\frac{1}{2}\)
c ) \(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{3}{4}\)
a)\(\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{-12}{19}=\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{7}{19}.\frac{-12}{7}=\frac{7}{19}.\left(\frac{8}{11}+\frac{3}{11}+-\frac{12}{7}\right)=\frac{7}{19}.\left(\frac{-5}{7}\right)=-\frac{5}{19}\)
b)\(\frac{-7}{25}.\frac{39}{-14}.\frac{50}{78}=\frac{\left(-7\right).39.50}{25.\left(-14\right).78}=\frac{\left(-7\right).3.13.2.5.5}{5.5.\left(-7\right).2.2.13.3}=\frac{1}{2}\)
c)\(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}=\frac{2}{7}+\frac{1}{2}=\frac{11}{14}\)
\(\frac{-3}{5}.\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
=\(\frac{-3}{5}.\)\(\frac{14}{7}\)
=\(-\frac{3}{5}.2\)
=-6/5
B=47/41.[12.(1+1/19-1/37-1/53)/3.(1+1/19-1/37-1/53):4.(1+1/17+1/19+1/2006)/5.(1+1/17+1/19+1/2006)].123/235
=47/41.[4:4/5].123/235
=47/41.5.123/235=3
C=63.10101.37-37.10101.63/1+2+3+...+2006
=0/1+2+3+...+2006=0
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2 ) B = \(1\frac{6}{41}.\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
B = 47/41 . ( 12/3 : 4/5 ) . 123/235
B = 47/41 . ( 4 : 4/5 ) . 123/235
B = 47/41 . 5 . 123/235
B = \(\frac{47.5.123}{41.235}\)
B = 3
1 ) A = \(\frac{636363.37-373737.63}{1+2+3+...+2006}\)
A = \(\frac{63.10101.37-37.10101.63}{1+2+3+...+2006}\)
A = \(\frac{0}{1+2+3+...+2006}\)
A = 0
a. \(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\)
\(=\frac{1}{3}(\frac{4}{5}+\frac{6}{5})-\frac{5}{3}\)
\(=\frac{1}{3}.2-\frac{5}{3}\)
\(=\frac{2}{3}-\frac{5}{3}\)
\(=-\frac{1}{1}\)
c. \(\frac{6}{7}.\frac{10}{9}+\frac{1}{7}.\frac{10}{9}-\frac{8}{9}\)
\(=\frac{10}{9}\left(\frac{6}{7}+\frac{1}{7}\right)-\frac{8}{9}\)
\(=\frac{10}{9}.1-\frac{9}{8}\)
\(=\frac{10}{9}-\frac{9}{8}\)
\(=-\frac{1}{72}\)
A = -10/3 + 19/6.7/5 - 19/3.1/10 + 19/10.4/3
A = 1/30.(-10.10 + 19.7 - 19.1 + 19.4)
A = 1/30.(-100 + 19.11 - 19)
A = 1/30.(-100 + 19.10)
A = 1/30.(-100 + 190)
A = 1/30.90
A = 3
bạn ơi ! cho mình hỏi 1/30 ở đâu vậy?