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1)
a) -(2+5) = -2 - 5 = -7
b) +(-3+6) = -3 + 6 = 3
c) (-50+3) = -50 + 3 = -47
d) -(-2+3) = 2 - 3 = -1
e) -(10-3) = -10 + 3 = -7
f) -(-3)-(-3+1) = 3 + 3 - 1 = 5
g) (-5)+(-2+10) = -5 - 2 + 10 = 3
2)
a) -50+120+(-150)-20+30
= -(50 + 20) + (120 + 30 - 150)
= -70
b) 265-70+(-65)-30+15
= (265 - 65) - (70 + 30) + 15
= 200 - 100 + 15 = 115
c) -17+185-183+(-85)-63
= (185 - 85) - (183 + 17) - 63
= 100 - 200 - 63 = -163
d) -30+60+(-170)-260+19
= -(170 + 30) - (260 - 60) + 19
= -200 - 200 + 19 = -381
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{2.6.10+4.12.20+...+20.60.100}{1.2.3+2.4.6+...+10.20.30}=\frac{2.6.10.1^3+2.6.10.2^3+...+2.6.10.10^3}{1.2.3.1^3+1.2.3.2^3+...+1.2.3.10^3}\)
\(=\frac{2.6.10.\left(1^3+2^3+...+10^3\right)}{1.2.3.\left(1^3+2^3+...+10^3\right)}=\frac{2.6.10}{1.2.3}=20\)
vậy M=20
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a) Đặt \(A=1+2+2^2+2^3+...+2^{100}\)
\(2A=2+2^2+2^3+...+2^{101}\)
\(2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...+2^{100}\right)\)
\(A=2^{101}-1< 2^{101}\)
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\(3\times24^{10}\)
\(=3\times\left(2^3\times3\right)^{10}\)
\(=3\times3^{10}\times\left(2^3\right)^{10}\)
\(=3^{11}\times2^{30}\)
\(=3^{11}\times\left(2^2\right)^{15}\)
\(=3^{11}\times4^{15}\)
Vì \(3^{11}\)<\(4^{15}\left(3;4;11;15\inℕ\right)\)
Nên \(3^{11}\times4^{15}\)< \(4^{15}\times4^{15}=4^{30}\)
Do đó : \(3\times24^{10}\)< \(4^{30}\)
Vậy \(2^{30}+3^{30}+4^{30}\)> \(3\times24^{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(10-\left[30-\left(3+2\right)^2\right]=10-30+5^2=-20+25=5.\)
\(10-\left[30\left(3+2\right)^2\right]\)
\(10-\left[30\cdot5^2\right]\)
\(10-750\)
\(=-740\)
10-(30(3+2)2]
= 10- [ 30 ( 9+4) ]
= 10 - ( 30 x 13 )
= 10- 390
= -740