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\(1024:64+\left[140:\left(38+\dfrac{2}{5}\right)-\dfrac{1}{3}\right]\\ =16+\left[140:\dfrac{192}{5}-\dfrac{1}{3}\right]\\ =16+\left[\dfrac{175}{48}-\dfrac{1}{3}\right]\\ =16+\dfrac{53}{16}\\ =\dfrac{309}{16}\)
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(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
B=[(45.79+45.21)]:90-5^2]:5+2^3 B=[(45.79+45.21):90-25]:5+8 B=[(45.(79+21):65]:13 B=[(45.100):65]:13 B=[4500:65]:13 B=4500:65:13
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\(1024\div2^5+140\div\left(38+5^2\right)-7^{23}\div7^{21}\)
\(=1024\div32+140\div\left(38+25\right)-7^{23}\div7^{21}\)
\(=32+140\div63-7^2\)
\(=32+\dfrac{20}{9}-49\)
\(=\dfrac{308}{9}-49\)
\(=-\dfrac{133}{9}\)
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a) 5^23 và 6 . 5^22
Ta có: 5^23 = 5^22 . 5
Vì 5 < 6 nên 5^23 < 6 . 5^22
b) 7 . 2^13 và 2^16
Ta có: 2^16 = 2^13 . 2^3 = 2^13 . 8
Vì 7 < 8 nên 7 . 2^13 < 2^16
c) 21^15 và 27^5 . 49^8
Ta có: 21^15 = (3.7)^15 = 3^15 . 7^15
27^5 . 49^8 = (3^3)^5 . (7^2)^8 = 3^15 . 7^16
Vì 7^15 < 7^16 nên 21^15 < 27^5 . 49^8
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a: =30-22=8
b: =10*(23+17)=10*40=400
c: =21*3-7*(-14)
=63+98=161
d: =-20-[10*10*5^2+16]
=-20-100*25-16
=-36-2500
=-2536
1024 : 2\(^4\) + 140 : ( 38 + 2\(^5\)) - 7\(^{23}\): 7\(^{21}\)
= 1024 : 16 + 140 : ( 38 + 32) - 7\(^{23-21}\)
= 64 + 140 : 70 - 7\(^2\)
= 64 + 2 - 49
= 17
Ủng hộ nha! :)
=17 nha