\(\frac{1}{3}M=\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{100}}\)

\(M-\frac{1}{3}M=\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+....+\left(\frac{1}{3^{99}}-\frac{1}{3^{99}}\right)+\frac{1}{3}-\frac{1}{3^{100}}\)

\(\frac{2}{3}M=\frac{1}{3}-\frac{1}{3^{100}}\)

Vậy \(M=\left(\frac{1}{3}-\frac{1}{3^{100}}\right):\frac{2}{3}=\frac{1}{2}-\frac{1}{2.3^{99}}<\frac{1}{2}\)

KL: M    < 1/2 (dpcm)

Câu 11

Câu 6

Câu 5

Câu 4

Câu 3

14C

454>578-987

487<758

4+5x5=29

11C