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12,6 + 102,47 - 178,51
= (126,34 + 102,47 ) - 178,51
= 228,81 - 178,51
= 50,3
k nhaa , minh dung 100%
Ta có: \(B=3+3^2+\cdots+3^{20}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\cdots+\left(3^{19}+3^{20}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+\cdots+3^{18}\left(3+3^2\right)\)
\(=12\left(1+3^2+\cdots+3^{18}\right)\) ⋮12
\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)
\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)
\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)
\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)
\(S=4\left(3^2+3^4+3^6+3^8\right)\)
\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4