Tính bằng cách thuận tiện nhất :
4 * 64,3*2,5 18,7*4,3+18,7*5,7 \(\frac{25}{141}\)*\(\frac{17}{11}\)-\(\frac{25}{141}\)*\(\frac{6}{11}\) 1+2+3+4+.......+2006
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a)=384:(25*4)=384:100=3,84.
b)37/25:5/4:74/25=37/25*4/5*25/74=37*4*25/5*25*74=2/5.
c)380:(38*2)=(380:38):2=10:2=5.
d)6/11:4/3:5/11:4/3=6/11*3/4*11/5*3/4=6*3*11*3/11*4*5*4=27/40.
k nha
a: \(=\dfrac{37}{25}\cdot\dfrac{25}{74}\cdot\dfrac{4}{5}=\dfrac{1}{2}\cdot\dfrac{4}{5}=\dfrac{2}{5}\)
b: \(=\dfrac{6}{11}\cdot\dfrac{11}{5}:\dfrac{16}{9}=\dfrac{6}{5}\cdot\dfrac{9}{16}=\dfrac{54}{80}=\dfrac{27}{40}\)
\(\frac{\frac{3}{11}-\frac{3}{13}+\frac{3}{17}-\frac{3}{19}}{\frac{4}{11}-\frac{4}{13}+\frac{4}{17}-\frac{4}{19}}=\frac{3\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}{4\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}=\frac{3}{4}\)
\(\frac{3.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}{4.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{17}-\frac{1}{19}\right)}\) = \(\frac{3}{4}\)
--------------------- Hok Tốt-----------------------------
a)\(\frac{21}{11}x\frac{22}{17}x\frac{68}{63}\)
=\(\frac{21x22x68}{11x17x63}\)
=\(\frac{21x11x2x17x4}{11x17x21x3}\)
=\(\frac{2x4}{3}\)
=\(\frac{8}{3}\)
b)\(\frac{5}{14}x\frac{7}{13}x\frac{26}{25}\)
=\(\frac{5x7x26}{14x13x25}\)
a, \(\frac{5}{11}\times\frac{7}{25}+\frac{15}{11}\times\frac{1}{5}\)
\(=\frac{5}{11}\times\frac{7}{25}+\frac{5}{11}\times\frac{3}{5}\)
\(=\frac{5}{11}\times\left(\frac{7}{25}+\frac{3}{5}\right)\)
\(=\frac{5}{11}\times\frac{22}{25}\)
\(=\frac{2}{5}\)
b, \(\frac{3}{7}\times\frac{25}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\frac{75}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\left(\frac{75}{19}-\frac{18}{19}\right)\)
\(=\frac{1}{7}\times3\)
\(=\frac{3}{7}\)
1) \(=\left(\frac{3}{5}+\frac{2}{5}\right).\frac{6}{11}\)
\(=1.\frac{6}{11}\)
\(=\frac{6}{11}\)
2)\(\frac{17}{25}.\left(\frac{11}{19}+\frac{6}{19}+\frac{2}{19}\right)\)
\(=\frac{17}{25}.1\)
\(=\frac{17}{25}\)
a.384:25:4=384:[25x4]=384:100=3.84
380:[38x2]=[380:38]x2=20
a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)
Phần b T2
c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)
a)
`40xx9,84xx0,25`
`=40xx0,25xx9,84`
`=10xx9,84`
`=98,4`
b)
`x+18,7=50,5:2,5`
`=>x+18,7=20,2`
`=>x=20,2-18,7`
`=>x=1,5`
a)40x9,84x0,25
=(40x0,25)x9,84
=10x9,84
=98,4
b)x+18,7=50,5:2,5
x+18,7=20,2
x=20,2-18,
x=1,5
a)
4 x 64,3 x 2,5 = 64,3 x (2,5 x 4) = 64,3 x 10 = 643
b)
18,7 x 4,3 + 18,7 x 5,7 = 18,7(4,3 + 5,7) = 18,7 x 10 = 187
c)
25/141 x 17/11 - 25/141 x 6/11 = 25/141 x (17/11 - 6/11) = 25/141 x 11/11 = 25/141 x 1 = 25/141
d)
số số hạng trong dãy : 2006-1+1 =2006(số)
1+2+3+4+...+2006 = (2006+1)+(2005+2)+(2004+3)+....+(1003+1004)
số cặp có tổng bằng 2007 là : 2006 : 2 = 1003(cặp)
Vậy 1+2+3+4+.....+2006 = 1003 x 2007 = 2013021