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\(\left(13\frac{4}{9}+2\frac{1}{9}\right)-3\frac{4}{9}\)
\(=\left(\frac{121}{9}+\frac{19}{9}\right)-\frac{31}{9}\)
\(=\frac{140}{9}-\frac{31}{9}\)
\(=\frac{109}{9}\)
\(1,25\div\frac{15}{20}+\left(25\%-\frac{5}{6}\right)\div4\frac{2}{3}\)
\(=1,25\div\frac{3}{4}+\left(\frac{25}{100}-\frac{5}{6}\right)\div\frac{14}{3}\)
\(=1,25\times\frac{4}{3}+\left(\frac{6}{24}-\frac{20}{24}\right)\div\frac{14}{3}\)
\(=\frac{5}{3}+\frac{-14}{24}\times\frac{14}{3}\)
\(=\frac{60}{36}+\frac{-98}{36}\)
\(=\frac{-38}{36}=\frac{-19}{18}\)
1+(-2)+3+(-4)+5+(-6)+...+21+(-22)=[1+(-2)]+[3+(-4)]+...+[21+(-22)]
-----------11 cặp----------------------
=(-1)+(-1)+...+(-1)=(-1).11=-11
a,\(\frac{7}{10}\cdot\frac{4}{9}+\frac{3}{10}\cdot\frac{4}{9}-1\frac{7}{9}\)
\(=\frac{14}{45}+\frac{2}{15}-\frac{16}{9}\)
\(=\frac{14}{45}+\frac{6}{45}-\frac{80}{45}\)
\(=\frac{-60}{45}=\frac{-4}{3}\)
b,\(\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{5}{4}-\frac{2}{3}\right)\cdot\left(-3\right)^2+\frac{5}{9}\cdot30\%\)
\(=\frac{-5}{6}+\frac{4}{9}\cdot\left(\frac{7}{12}\right)\cdot9+\frac{5}{9}\cdot\frac{3}{10}\)
\(=\frac{-5}{6}+\frac{7}{3}+\frac{1}{6}\)
\(=\frac{-5}{6}+\frac{14}{6}+\frac{1}{6}\)
=\(=\frac{10}{6}=\frac{5}{3}\)
C =[125.(73-75)]:(-25)2
=125.(-2):(-25).2
=-250:(-50)
=5
E=125.9.(-4).(-8).25.7
=125.(-8).25.(-4).7.9
=-1000.(-100).63
=6300000
C = ( 125 . 73 - 125 . 75 ) : ( -50)
C = 125 .( -2 ) : (-50)
C = -250 : -50
C = 50
G = ( -3 )2+( -5)3: l -5 l
G = 9 + ( -125) : 5
G = 9 + ( -25 )
G= ( -16 )
E = 125 .9.*( -4 ) . ( -8) .25.7
E = 25.5 .( -4 ) .9 .2. ( -4 ) .25 .7
E =( -100 ).10.( -100).63
E = 10000.630
E = 630000
**** mha
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
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