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Đặt: \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2013}\right)\)
\(=\frac{1}{2}.\frac{2012}{2013}\)
\(=\frac{1006}{2013}\)
58 x 42 + 32 x 8 + 5 x16
= (2x29)(21x2) +(4x8)x8 + 5x(4x4)
= 4 x 609 + 4 x 64 + 4x20
= 4( 609+ 64+ 20)
= 4 x 693
= 2772
Đặt A=\(\frac{1}{3}.5+\frac{1}{5}.7+...+\frac{1}{97}.99\)
=>A=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
=>2A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
=>2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
=>2A=\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
=>A=\(\frac{32}{99}:2=\frac{32}{99}.\frac{1}{2}=\frac{32}{198}=\frac{16}{99}\)
Gọi số đó là a
Ta có:
\(\left\{\left[\left(8a+1\right)\frac{4}{7}+\frac{1}{2}\right]-\frac{1}{2}\right\}:3+\frac{1}{2}=0,6\)
\(\left\{\left[\left(8a+1\right)\frac{4}{7}+\frac{1}{2}\right]-\frac{1}{2}\right\}:3=\frac{1}{5}\)
\(\left[\left(8a+1\right)\frac{4}{7}+\frac{1}{2}\right]-\frac{1}{2}=\frac{1}{15}\)
\(\left(8a+1\right)\frac{4}{7}+\frac{1}{2}=\frac{17}{30}\)
\(\left(8a+1\right)\frac{4}{7}=\frac{1}{15}\)
\(8a+1=\frac{7}{45}\)
\(8a=\frac{-38}{45}\)
\(a=\frac{-19}{180}\)
\(\left\{\left[\left(8a+1\right)\frac{4}{7}+\frac{1}{2}\right]-\frac{1}{2}\right\}:3=\frac{1}{5}\)
2 x 3 x 5 + 3 + 5 + 11 = 49
Chắc thế
HT
:)))
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