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(9-8-7-6-5-4-3-2-1).(500.9-250.18)
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CM
27 tháng 3 2017
a) 815 – 23 – 77 +185 = (815 + 185) – (23+77) = 1000 – 100 = 900
b) 3145 – 246 + 2347 – 145 + 4246 – 347
= (3145-145) + (4246 – 246) + (2347 – 347)
= 3000 + 4000 + 2000 = 9000
c) (2018:1 – 2018.1):(2018.2008+2018.2002) = 0 : (2018.2008+2018.2002) = 0
d) (9 – 8 – 7 – 6 – … – 2 – 1).(500.9 – 250.18)
= (9 – 8 – 7 – 6 – … – 2 – 1).(500.9 – 250.2.9)
= (9 – 8 – 7 – 6 – … – 2 – 1).(500.9 – 500.9)
= (9 – 8 – 7 – 6 – … – 2 – 1).0 = 0
14 tháng 10 2021
\(a,=27\cdot20+73\cdot20-27=20\left(27+73\right)-27\\ =20\cdot100-27=2000-27=1973\\ b,=42\cdot50-38\cdot50-50\cdot2=50\left(42-38-2\right)=50\cdot2=100\\ e,=254\left(58+21\cdot2\right)=254\cdot100=25400\\ c,=53\left(12+172-84\right)=53\cdot100=5300\\ d,=24\left(31+42+27\right)=24\cdot100=2400\\ f,=\left(9-8-...-1\right)\left(500\cdot9-500\cdot9\right)=\left(9-8-...-1\right)\cdot0=0\)
11 tháng 3 2023
1/2 + 2/3 - 3/4 * 4/5 * 5/6 * 6/7 * 7/8 * 8/9 *...* 29/30 : 6/10
= 1/2 + 2/3 - 1/10 : 6/10
= 1/2 + 2/3 - 1/6
= 1
NQ
4
LD
31 tháng 12 2019
=(1-2)+(3-4)+...+(51-52)+53
= -1 + -1 + ... + -1 + 53
= -1 x (52:2)+53
= -26 + 53
=27
DD
8 tháng 2 2023
`5`
`a, -7/21 +(1+1/3)`
`=-7/21 + ( 3/3 + 1/3)`
`=-7/21+ 4/3`
`=-7/21+ 28/21`
`= 21/21`
`=1`
`b, 2/15 + ( 5/9 + (-6)/9)`
`= 2/15 + (-1/9)`
`= 1/45`
`c, (9-1/5+3/12) +(-3/4)`
`= ( 45/5-1/5 + 3/12)+(-3/4)`
`= ( 44/5 + 3/12)+(-3/4)`
`= 9,05 +(-0,75)`
`=8,3`
`6`
`x+7/8 =13/12`
`=>x= 13/12 -7/8`
`=>x=5/24`
`-------`
`-(-6)/12 -x=9/48`
`=> 6/12 -x=9/48`
`=>x= 6/12-9/48`
`=>x=5/16`
`---------`
`x+4/6 =5/25 -(-7)/15`
`=>x+4/6 =1/5 + 7/15`
`=> x+ 4/6=10/15`
`=>x=10/15 -4/6`
`=>x=0`
`----------`
`x+4/5 = 6/20 -(-7)/3`
`=>x+4/5 = 6/20 +7/3`
`=>x+4/5 = 79/30`
`=>x=79/30 -4/5`
`=>x= 79/30-24/30`
`=>x= 55/30`
`=>x= 11/6`
VN
8 tháng 2 2023
\(5)\)
\(A=\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)
\(A=\dfrac{-7}{21}+\dfrac{4}{3}\)
\(A=\dfrac{-7}{21}+\dfrac{28}{21}\)
\(A=1\)
\(--------------\)
\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+\dfrac{-6}{9}\right)\)
\(B=\dfrac{2}{15}+\dfrac{-1}{9}\)
\(B=\dfrac{18}{135}+\dfrac{-15}{135}\)
\(B=\dfrac{1}{45}\)
\(------------\)
\(C=9-\dfrac{1}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)
\(C=\dfrac{44}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)
\(C=\dfrac{528}{60}+\dfrac{15}{60}+\dfrac{-3}{4}\)
\(C=\dfrac{181}{20}+\dfrac{-3}{4}\)
\(C=\dfrac{181}{20}+\dfrac{-15}{20}\)
\(C=\dfrac{83}{10}\)
\(6)\)
\(a)\) \(x+\dfrac{7}{8}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}-\dfrac{7}{8}\)
\(x=\dfrac{104}{96}-\dfrac{84}{96}\)
\(x=\dfrac{5}{24}\)
\(b)\) \(\dfrac{-6}{12}-x=\dfrac{9}{48}\)
\(\dfrac{-1}{2}-x=\dfrac{3}{16}\)
\(x=\dfrac{-1}{2}-\dfrac{3}{16}\)
\(x=\dfrac{-8}{16}-\dfrac{3}{16}\)
\(x=\dfrac{-11}{16}\)
\(c)\) \(x+\dfrac{4}{6}=\dfrac{5}{25}-\left(-\dfrac{7}{15}\right)\)
\(x+\dfrac{4}{6}=\dfrac{5}{25}+\dfrac{7}{15}\)
\(x+\dfrac{4}{6}=\dfrac{75}{375}+\dfrac{105}{375}\)
\(x+\dfrac{4}{6}=\dfrac{12}{25}\)
\(x=\dfrac{12}{25}-\dfrac{4}{6}\)
\(x=\dfrac{72}{150}-\dfrac{100}{150}\)
\(x=\dfrac{-14}{75}\)
\(d)\) \(x+\dfrac{4}{5}=\dfrac{6}{20}-\left(-\dfrac{7}{3}\right)\)
\(x+\dfrac{4}{5}=\dfrac{6}{20}+\dfrac{7}{3}\)
\(x+\dfrac{4}{5}=\dfrac{18}{60}+\dfrac{140}{60}\)
\(x+\dfrac{4}{5}=\dfrac{79}{30}\)
\(x=\dfrac{79}{30}-\dfrac{4}{5}\)
\(x=\dfrac{79}{30}-\dfrac{24}{30}\)
\(x=\dfrac{11}{6}\)
NL
4
PV
16 tháng 3 2017
13 - 12 + 11 + 10 - 9 + 8 - 7 - 6 + 5 - 4 + 3 + 2 - 1
= 1 + 11 - 1 + 1 + 11 - 7 + 1
= 12 - 1 + 1 + 11 - 7 + 1
= 11 + 1 + 11 - 7 + 1
= 12 + 11 - 7 + 1
= 23 - 7 + 1
= 16 + 1
=17
LM
1
AH
Akai Haruma
Giáo viên
31 tháng 12 2023
Lời giải:
$A=(1+2-3-4)+(5+6-7-8)+(9+10-11-12)+....+(297+298-299-300)+301+302-303$
$=(-4)+(-4)+(-4)+....+(-4)+300$
Số lần xuất hiện của $-4$ là:
$[(300-1):1+1]:4=75$
$A=(-4),75+300=0$
\(=\left(9-8-...-2-1\right)\left(500\cdot9-500\cdot9\right)=0\)
0