tính nhanh:a]2/5:2/7*3/7:2/3+1999
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5 tháng 12 2021
a)184 x 72 + 184 x 27 + 184
=184 x 72 + 184 x 27 + 184 x 1
= 184 x (72 + 27 + 1 )
=184 x 100=18400
c)1999 x 346 + 1999 + 653 x 1999
=1999 x 346 + 1999 + 653 x 1999 x 1
= 1999 x ( 346 + 653 + 1 )
= 1999 x 1000=1999000
e)23 x 5 x 7 x 4
= ( 23 x 7) x( 5 x 4)
= 161 x 20= 3220
b)1996 x 1257 – 1996 – 256 x 1996
=1996 x 1257 – 1996 – 256 x 1996 x 1
= 1996 x ( 1257 - 256 - 1 )
= 1996 x 1000= 1996000
MC
2
NH
27 tháng 2 2023
a)
`1/3+3/4+2/3+1/4`
`=1/3+2/3+3/4+1/4`
`=1+1`
`=2`
b)
`3/4+3/5+2/8+4/10`
`=3/4+2/8+3/5+4/10`
`=6/8+2/8+6/10+4/10`
`=1+1`
`=2`
c)
`1/10+2/10+3/10+4/10+5/10+6/10+7/10+8/10+9/10`
`=1/10+9/10+2/10+8/10+3/10+7/10+6/10+4/10+5/10`
`=1+1+1+1+5/10`
`=4+5/10`
`=40/10+5/10`
`=45/10=9/2`
27 tháng 2 2023
a: =1/3+2/3+3/4+1/4
=1+1
=2
b: =3/4+1/4+3/5+2/5
=1+1
=2
c: =(1+2+3+4+5+6+7+8+9)/10
=45/10
=9/2
T
23 tháng 8 2023
\(a,13\dfrac{3}{5}-\left(8\dfrac{3}{5}-4\dfrac{3}{4}\right)\)
\(=\dfrac{68}{5}-\dfrac{43}{5}+\dfrac{19}{4}\)
\(=5+\dfrac{19}{4}\)
\(=\dfrac{20}{4}+\dfrac{19}{4}=\dfrac{39}{4}\)
\(------\)
\(b,11\dfrac{1}{4}-\left(2\dfrac{5}{7}+5\dfrac{1}{4}\right)\)
\(=\dfrac{45}{4}-\dfrac{19}{7}-\dfrac{21}{4}\)
\(=\left(\dfrac{45}{4}-\dfrac{21}{4}\right)-\dfrac{19}{7}\)
\(=6-\dfrac{19}{7}\)
\(=\dfrac{42}{7}-\dfrac{19}{7}=\dfrac{23}{7}\)
17 tháng 2 2022
a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)
b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)
d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)
VT
1
29 tháng 8 2021
Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{8}{9}\)
\(=\dfrac{1}{9}\)
28 tháng 7 2019
a.\(\frac{3}{5}\)\(.\)\(\frac{1195}{1999}\)+ \(\frac{4}{1999}\)\(.\)\(\frac{3}{5}\)= \(\frac{3}{5}.\left(\frac{1195}{1999}+\frac{4}{1999}\right)\)= \(\frac{3}{5}.1\)= \(\frac{3}{5}\)
28 tháng 7 2019
b. \(\frac{2}{3}\)\(:\)\(\frac{5}{7}\)\(.\)\(\frac{5}{7}\)\(:\)\(\frac{2}{3}\)\(+1934\)= \(\frac{2}{3}\).\(.\)\(\frac{7}{5}\)\(.\)\(\frac{5}{7}\)\(.\)\(\frac{3}{2}\)\(+\)\(1934\)= \(1\)\(+\)\(1934\)= \(1935\)
3 tháng 7 2023
câu b bài 2 thiếu nha e
B1
a, 638 +780 x 5 - 369 : 9 = 4497
b, ( 273 + 485 ) x16 - 483 :3 x4 = 11484
B2
a, 325 x 6 + 6 x 560 + 115 = 5425
B3
a, x = 1020
b, x = 36
B4
26 nha
H9
HT.Phong (9A5)
CTVHS
17 tháng 1
a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)
\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)
\(=1-\dfrac{29}{15}\)
\(=\dfrac{15-29}{15}\)
\(=-\dfrac{14}{15}\)
b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)
\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{7}{9}\cdot1\)
\(=\dfrac{7}{9}\)
c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{25}{63}\)
d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)
DD
Đoàn Đức Hà
Giáo viên
6 tháng 2 2021
a) \(A=1+\left(-3\right)+5+\left(-7\right)+...+\left(-1999\right)+2001\)
Số số hạng của tổng trên là: \(\frac{2001-1}{2}+1=1001\).
\(A=\left[1+\left(-3\right)\right]+\left[5+\left(-7\right)\right]+...+\left[1997+\left(-1999\right)\right]+2001\)
\(A=-2.500+2001\)
\(A=1001\)
DD
Đoàn Đức Hà
Giáo viên
6 tháng 2 2021
b) \(1+\left(-2\right)+\left(-3\right)+4+5+\left(-6\right)+\left(-7\right)+8+...+1997+\left(-1998\right)+\left(-1999\right)+2000\)
\(=\left\{\left[1+\left(-2\right)\right]+\left[\left(-3\right)+4\right]\right\}+...+\left\{\left[1997+\left(-1998\right)\right]+\left[\left(-1999\right)+2000\right]\right\}\)
\(=\left(-1+1\right)+\left(-1+1\right)+...+\left(-1+1\right)\)
\(=0+0+...+0=0\)
\(\frac{19999}{10}\) =