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a) \(26+173+74+27\)
\(=\left(26+74\right)+\left(173+27\right)\)
\(=100+200\)
\(=300\)
b) \(75\cdot37+89\cdot46+75\cdot52-89\cdot21\)
\(=75\cdot\left(37+52\right)+89\cdot\left(46-21\right)\)
\(=75\cdot89+89\cdot25\)
\(=89\cdot\left(75+25\right)\)
\(=89\cdot100\)
\(=8900\)
c) \(2^7:2^2+5^4:5^3\cdot2^4-3\cdot2^5\)
\(=2^{7-2}+5^{4-3}\cdot2^4-3\cdot2^5\)
\(=2^5+5\cdot2^4-3\cdot2^5\)
\(=2^4\cdot\left(2+5-3\cdot2\right)\)
\(=2^4\cdot\left(7-6\right)\)
\(=2^4\)
\(=16\)
d) \(100:\left\{250:\left[450-\left(4\cdot5^3-2^2\cdot25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(4\cdot5^3-4\cdot5^2\right)\right]\right\}\)
\(=100:\left[250:\left(450-4\cdot5^2\cdot4\right)\right]\)
\(=100:\left[250:\left(450-400\right)\right]\)
\(=100:\left(250:50\right)\)
\(=100:5\)
\(=20\)
a) 20 : 2 2 − 5 9 : 5 8 = 5 − 5 = 0
b) ( 5 19 : 5 17 − 4 ) : 7 0 = ( 5 2 − 4 ) : 1 = 21
c) 295 − ( 31 − 2 2 .5 ) 2 = 295 − 11 2 = 174
\(a)\)
\(\left(-31\right)+\left(50-19\right)-\left(150-31\right)\)
\(=\left(-31\right)+50-19-150+31\)
\(=\left(-150\right)-19\)
\(=-169\)
\(b)\)
\(25.\left(45-17\right)-45.\left(25-17\right)\)
\(=25.45-25.17-45.25+45.17\)
\(=0\)
\(c)\)
\(\frac{-1}{12}+\frac{4}{3}=\frac{5}{4}\)
\(d)\)
\(3+\frac{-5}{20}+\frac{30}{75}+\frac{-7}{4}\)
\(=\left(\frac{3}{5}+\frac{30}{75}\right)-\left(\frac{5}{20}+\frac{7}{4}\right)\)
\(=1-2\)
\(=-1\)
a) (-31)+(50-19)-(150-31)
= (-31)+50+(-19)-150+(-31)
= (-31)+50-150+(-19)-(-31)
= (-31)+(-100)+12
= -119
b) 25(45-17)-45(25-17)
= 25.45-25.17-45.25-45.17
= 25(45-45)-25(17-17)
= 0
c) -1/12 + 4/3
= -1/12 + 16/12
= 15/12
= 5/4
d) 3/5+(-5)/20+30/75+(-7)/4
= 45/75+30/75+(-5)/20+(-35)/20
= 1+(-2)
= -1
a) \(35.26+35.74\)
\(=35\left(26+74\right)\)
\(=35.100\)
\(=3500\)
b) \(6^7:6^5+3.3^2\)
\(=6^2+3^3\)
\(=36+27\)
\(=63\)
a,a) 5.(-8).2.(-7) =560
b)-175+49+275-149=0
c)34.15+15.(-74)=-600
a) 5 . ( -8 ) . 2 . ( -7 )
= [ 5 . 2 ] . [ -8 . ( -7 ) ]
= 10 . 56
= 560
b) -175 + 49 + 275 - 149
= [ -175 + 275 [ + [ 49 - 149 ]
= 100 + ( -100 )
= 0
c) 34 . 15 + 15 . ( -74 )
= 15 . [ 34 + ( -74 ) ]
= 15 . ( -40 )
= -600
d) 1 - 2 + 3 - 4 +.....+ 2009 - 2010 + 2011
Dãy trên có số số hạng là:
( 1011 - 1 ) : 1 + 1 = 1011 ( số hạng )
Ta ghép mỗi bộ 2 số vậy có 505 bộ lẻ số cuối cùng.
Ta có:
1 - 2 + 3 - 4 +....+ 2009 - 2010 + 2011
= ( 1 - 2 ) + ( 3 - 4 ) + ..... + ( 2009 - 2010 ) + 2011
= ( -1 ) + ( -1 ) + ....+ ( -1 ) + 2011
Dãy trên có 505 số ( -1 ) và lẻ 2011
Vậy tổng trên là:
505 . ( -1 ) + 2011 = 1506
a) \(3.5^2+15.2^2-26\div2\)
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) \(5^3.2-100\div4+2^3.5\)
= 125.2 - 25 + 8.5
= 250 - 25 + 40
= 225 + 40
= 265
c)\(6^2\div9+50.2-3^3.33\)
= 36 : 9 + 100 - 9.33
= 4 + 100 - 297
= 104 - 297
= -193
d)\(3^2.5+2^3.10-81\div3\)
= 9.5 + 8.10 - 27
= 45 + 80 - 27
= 125 - 27
= 98
e) \(5^{13}\div5^{10}-25.2^2\)
= 53 - 25.4
= 125 - 100
= 25
f) \(20\div2^2+5^9\div5^8\)
= 20 : 4 + 5
= 5 + 5
= 10
\(a.36+15+64\\ =\left(36+64\right)+15\\ =100+15\\ =115\\ b.12\cdot36+12\cdot64\\ =12\cdot\left(36+64\right)\\ =12\cdot100\\ =1200\\ c.\left(3^2+2^3\cdot5\right):7\\ =\left(9+8\cdot5\right):7\\ =\left(9+40\right):7\\ =49:7\\ =7\)\(\)
a) \(26+150+74=\left(26+74\right)+150=100+150=250\).
b) \(35.46+54.35=35.\left(46+54\right)=35.100=3500\).
c) \(\left(-31\right)+\left[20-\left(7-4\right)^2\right]=\left(-31\right)+20-3^2=\left[\left(-31\right)-9\right]+20=-40+20=-20\)