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\(a,-\left(793-2015\right)+\left(-2015-1207\right)\)
\(=-793+2015-2015-1207\)
\(=\left(-793-1207\right)+\left(2015-2015\right)\)
\(=-2000+0\)
\(=-2000\)
\(b,3^2+\left\{-54:\left[\left(-2\right)^3+7.|-2|\right].\left(-2\right)^2\right\}\)
\(=9+\left\{-54:\left[-8+7.2\right].4\right\}\)
\(=9+\left[-54:6.4\right]\)
\(=9+\left(-36\right)\)
\(=9-36\)
\(=-27\)
\(-3^2+\left\{-54:\left[\left(-2\right)^3+7.|-2|\right]\right\}.\left(-2\right)^2\)'
=\(9+\left\{-54:\left[\left(-8\right)+7.2\right]\right\}.4\)
\(=9+\left\{-54:\left[\left(-8\right)+14\right]\right\}.4\)
\(=9+\left\{-54:6\right\}.4\)
\(=9+\left(-9\right).4\)
\(=9+\left(-36\right)\)
\(=-27\)
chúc bạn học tốt
\(A=2^0+2^1+2^2+...+2^{2010}\)
\(\Rightarrow2A=1+2^2+2^3+...+2^{2011}\)
\(\Rightarrow2A-A=2011-1\)\(\Rightarrow A=B\)
a, \(A=3+3^2+3^3+...+3^{54}.\)
\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{52}+3^{53}+3^{54}^{ }3\right)\)
\(\Rightarrow3.\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{52}\left(1+3+3^2\right)\)
\(\Rightarrow13.\left(3+3^4+...+3^{52}\right)\)
\(\Rightarrow A⋮13\)
b, tương tự
b) 3^2 . [(5^2 - 3 ) : 11 ] - 2^4 + 2.10^3
= 9 . [(25 - 3 ) : 11 ] - 16 + 2.1000
= 9 . [22 : 11 ] - 16 + 2000
= 9 . 2 - 16 + 2000
= 18 - 16 + 2000
= 2 + 2000
= 2002
(72005 + 72004) : 72004
= 72005 : 72004 + 72004 : 72004
= 72005 - 2004 + 1
= 71 + 1
= 7 + 1
= 8
a) ( 3^5 . 3^7 ) : 3^10 + 5.2^4 - 7^3 : 7
= 3^10 : 3^10 + 80 - 7^2
= 1 + 80 - 49
= 32
a) 46-(-54)+78-(50+78)
=46+54+78-50-78
=(46+54)+(78-78)-50
=100+0-50
=50
b) 125.5.(-2).(-8).(-6)
=[125.(-8)].[5.(-2)].(-8)
=-200.(-10).(-8)
=2000.(-8)
=-16000
c) (-2)3.4+3.(-4)2-18
=-8.4+3.16-18
=-32+48-18
=16-18
=-2
\(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{22.3^{28}-3^{28}.2}{2^2.3^{28}}=\frac{3^{28}.20}{3^{28}.4}=\frac{20}{4}=5\)
Mk ko hiểu cách tl của bn Nguyễn Ngọc Quý ak.Bn có thể giải thích rõ hơn ko?
Gọi 1+2+22+....+2100 là A
Ta có:
A=1+2+22+....+2100
2A=2+22+23+...+2101
2A-A=(2+22+23+...+2101)-(1+2+22+23+...+2100)
A=2101-1