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1/1*4 + 1/4*7 + 1/7*10 + ... + 1/97*100
= 1/3(3/1*4 + 3/4*7 + 3/7*10 + ... + 3/97*100)
= 1/3(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100)
= 1/3(1 - 1/100)
= 1/3*99/100
= 33/100
\(=\frac{1}{3}x\left(\frac{3}{1x4}+\frac{3}{4x7}+...+\frac{3}{77x80}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}x\left(\frac{1}{1}-\frac{1}{80}\right)\)
\(=\frac{1}{3}\times\frac{79.}{80}\)
\(=\frac{79}{240}\)
Tk giúp mk nha cảm ơn !!
S=1/1-1/4+1/4+1/7-1/7+1/10+...+1/100-1/103
S=1/1-1/103
S=102/103
Vì 102/103<1 nên S<1
1. \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}\)
\(=\frac{42}{43}\)
2. Đặt \(A=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2.\left(1-\frac{1}{10}\right)\)
\(=2.\frac{9}{10}\)
\(=\frac{9}{5}\)
Ủng hộ mk nha !!! ^_^
1) 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/40×43
= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/40 - 1/43
= 1 - 1/43
= 42/43
2) 2/2 + 2/6 + 2/12 + ... + 2/90
= 2 × (1/2 + 1/6 + 1/12 + ... + 1/90)
= 2 × (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/9×10)
= 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)
= 2 × (1 - 1/10)
= 2 × 9/10
= 9/5
\(D=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(D=1-\frac{1}{100}\)
\(D=\frac{99}{100}\)
\(D=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(D=\frac{2}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\left(1-\frac{1}{100}\right)\)
\(D=\frac{2}{3}\cdot\frac{99}{100}=\frac{33}{50}\)