\(\dfrac{15}{28} : \dfrac57\)
A = \(\dfrac{28}{25}\) + \(\dfrac{9}{10}\) - \(\dfrac{11}{15}\) + \(\dfrac{13}{21}\) - \(\dfrac{15}{28}\) + \(\dfrac{17}{26}\) - ... + \(\dfrac{197}{4851}\) - \(\dfrac{199}{4950}\)
ét ô ét
\((\)\(\dfrac57 - \dfrac1{14}\)\()\) x \(7 = \)
\(\dfrac{9}{14}\times7=\dfrac{63}{14}=\dfrac{9}{2}\)
\(\left(\dfrac{5}{7}-\dfrac{1}{14}\right)\times7=\dfrac{9}{14}\times7=\dfrac{9}{2}\)
tính
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
b)\(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}+\dfrac{29}{42}:\dfrac{1}{28}-8\)
c)\(-1\dfrac{5}{7}.15+\dfrac{2}{7}\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
c; C = -1\(\dfrac{5}{7}\).15 + \(\dfrac{2}{7}\)(-15) + (-105).(\(\dfrac{2}{3}\) - \(\dfrac{4}{5}\) + \(\dfrac{1}{7}\))
C = - 15.(- 1 - \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\) + \(1\))
C = -15.[(1 - 1) - (\(\dfrac{5}{7}\) - \(\dfrac{2}{7}\)) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15.[0 - \(\dfrac{3}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15 . [- \(\dfrac{45}{105}\) + \(\dfrac{490}{105}\) - \(\dfrac{588}{105}\)]
C = -15. [ \(\dfrac{445}{105}\) - \(\dfrac{588}{105}\)]
C = - 15.(- \(\dfrac{143}{105}\))
C = \(\dfrac{143}{7}\)
\(\dfrac{11}{9}\) - \(\dfrac{5}{6}\) : 15
\(\dfrac{3}{16\:}\)+ \(\dfrac{ }{ }\)\(\dfrac{7}{30}\) x \(\dfrac{15}{28}\)
11/9-5/6:15
=11/9-5/6.1/15
=11/9-1/18
=7/6
3/16+(-7/30).15/28
=3/16+(-1/8)
=3/16+(-2/16)
=1/16
\(\dfrac{11}{9}-\dfrac{5}{6}:15=\dfrac{11}{9}-\dfrac{5}{6}\times\dfrac{1}{15}=\dfrac{11}{9}-\dfrac{5}{90}=\dfrac{110}{90}-\dfrac{5}{90}=\dfrac{105}{90}=\dfrac{7}{6}\)
\(\dfrac{3}{16}+\dfrac{-7}{30}\times\dfrac{15}{28}=\dfrac{3}{16}+\left(-\dfrac{1}{8}\right)=\dfrac{3}{16}-\dfrac{1}{8}=\dfrac{3}{16}-\dfrac{2}{16}=\dfrac{1}{16}\)
Bài 1.Tính
a)\(\dfrac{4}{28}\)+\(\dfrac{12}{36}\) b)\(\dfrac{-12}{18}\)+\(\dfrac{15}{-21}\) c)\(\dfrac{14}{28}\)-\(\dfrac{-16}{32}\)-\(\dfrac{17}{51}\)
\(a,\dfrac{1}{7}+\dfrac{1}{3}=\dfrac{3}{21}+\dfrac{7}{21}=\dfrac{10}{21}\\ b,\dfrac{-2}{3}+\dfrac{-5}{7}=\dfrac{-14+\left(-15\right)}{21}=\dfrac{-29}{21}\\ c,\dfrac{1}{2}-\dfrac{-1}{2}-\dfrac{1}{3}=\dfrac{3+3-2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(a.\dfrac{4}{28}+\dfrac{12}{36}=\dfrac{1}{7}+\dfrac{1}{3}=\dfrac{3}{21}+\dfrac{7}{21}=\dfrac{10}{21}\\ b.\dfrac{-12}{18}+\dfrac{-15}{21}=\dfrac{-2}{3}+\dfrac{-5}{7}=\dfrac{-14}{21}+\dfrac{-15}{21}=\dfrac{-29}{21}\\ c.\dfrac{14}{28}+\dfrac{16}{32}-\dfrac{17}{51}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{17}{51}=1-\dfrac{17}{51}=\dfrac{2}{3}\)
\(\dfrac{4}{28}+\dfrac{12}{36}=\dfrac{144}{1008}+\dfrac{336}{1008}=\\ \dfrac{480}{1008}=\dfrac{10}{21}\)
Câu b và c em làm tương tự.
C= 2-\(\dfrac{5}{3}\)+\(\dfrac{7}{6}\)-\(\dfrac{9}{10}\)+\(\dfrac{11}{15}\)-\(\dfrac{13}{21}\)+\(\dfrac{15}{28}\)-\(\dfrac{17}{36}\)+\(\dfrac{19}{45}\)
tính C
\(=2-\left(\dfrac{5}{3}-\dfrac{7}{6}+\dfrac{9}{10}-...-\dfrac{19}{45}\right)\)
\(=2-2\left(\dfrac{5}{6}-\dfrac{7}{12}+\dfrac{9}{20}-...-\dfrac{19}{90}\right)\)
\(=2-2\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{5}-...-\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=2-2\cdot\dfrac{4}{10}=2-\dfrac{8}{10}=2-\dfrac{4}{5}=\dfrac{6}{5}\)
\(\dfrac{15}{x}\)+\(\dfrac{3}{4}\)=\(\dfrac{7}{28}\)
giúp với
\(\Rightarrow\dfrac{15}{x}=\dfrac{7}{28}-\dfrac{3}{4}\\ \Rightarrow\dfrac{15}{x}=\dfrac{-2}{4}\\ \Rightarrow15.4=x.\left(-2\right)=60\\ \Rightarrow x=60:\left(-2\right)=-30\\ \Rightarrow x=-30\)
Cho \(A=\dfrac{13}{25}+\dfrac{9}{10}-\dfrac{11}{15}+\dfrac{13}{21}-\dfrac{15}{28}+\dfrac{17}{36}-...+\dfrac{197}{4851}-\dfrac{199}{4950}\)
Chứng minh \(A>\dfrac{9}{10}\)
\(\dfrac{-28}{35}\)=\(\dfrac{16}{x}\)
\(\dfrac{x+7}{15}\)=\(\dfrac{-24}{36}\)
giúp mình
a: \(\Leftrightarrow\dfrac{16}{x}=\dfrac{-4}{5}\)
hay x=-20
b: =>x+7=-10
hay x=-17
\(-\dfrac{28}{35}=\dfrac{16}{x}\Leftrightarrow-\dfrac{4}{5}=\dfrac{16}{x}\)
\(\Rightarrow-4x=5\cdot16\)
\(\Rightarrow-4x=80\)
\(\Rightarrow x=-20\)
\(\dfrac{x+7}{15}=-\dfrac{24}{36}\Leftrightarrow\)\(\dfrac{x+7}{15}=-\dfrac{2}{3}\)
\(\Rightarrow3\left(x+7\right)=-2\cdot15\)
\(x+7=-30\div3\)
x+7=-10
x=-10-7
x=-17