4 .{ 32 [ (52 + 23) : 11] - 26} + 2010
[ ( 25 - 22 . 3) + (32 . 4 + 16)] :5
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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
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37
=(1+37)x37:2
=703
1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*0*20*21*22*23*24*25*26*27*28*29*30*31*32=0
A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
C = 2-3 + (52)3.5-3 + 4-3.16 - 2.32 - 105.(\(\dfrac{24}{51}\))0
C = \(\dfrac{1}{8}\) + 56.5-3 + 4-3.42 - 2.9 - 105.1
C = \(\dfrac{1}{8}\) + 53 + \(\dfrac{1}{4}\) - 18 - 105
C = (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\)) - (105 - 125 + 18)
C = \(\dfrac{3}{8}\) - (-20 + 18)
C = \(\dfrac{3}{8}\) + 2
C = \(\dfrac{19}{8}\)
3: =231+26-209-26=22
4: =19(25+30+45)=19x100=1900
5: =-50+19+143+79-25-48
=-75+95+98
=98+20=118
\(4.\left\{3^2\left[\left(5^2+2^3\right):11\right]-26\right\}+2010=4.\left\{3^2\left[\left(25+8\right):11\right]-26\right\}+2010\)
\(4\left\{3^2\left[33:11\right]-26\right\}+2010=4.\left\{3^2.3-26\right\}+2010=4\left\{27-26\right\}+2010\)
\(=4.1+2010=2014\)
\(\left[\left(25-2^2.3\right)+\left(3^2.4+16\right)\right]:5=\left[\left(25-4.3\right)+\left(9.4+16\right)\right]:5\)
\(=\left[\left(25-12\right)+\left(36+16\right)\right]:5=\left[13+52\right]:5=65:5=13\)