a) \(x+1+x+2+x+3+x+\frac{1}{4}+x+100=7450\)
\(\Leftrightarrow5x+\frac{425}{4}=7540\)
\(\Leftrightarrow x=\frac{5947}{4}\)
Vậy...
b) \(M=3+3^2+3^3+3^4+...+3^{99}+3^{100}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)\)
\(=4\left(3+3^3+...+3^{99}\right)⋮4\)
Ta lại có:
\(M=3+3^2+3^3+3^4+...+3^{99}+3^{100}\)
\(=\left(3+3^3\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)
\(=\left(3+3^2\right)+3\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)
\(=12\left(1+3+...+3^{98}\right)⋮12\)
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