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1) A = \(\frac{-15}{19}.\frac{23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\frac{-23}{37}+\frac{14}{37}.\frac{15}{19}=\frac{15}{19}.\left(\frac{-23}{37}+\frac{14}{37}\right)=\frac{15}{19}.\frac{-9}{37}=\frac{-135}{703}\)
A = 1/4 + 1/28 + 1/70 +...+ 1/9700
A = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/97.100
3A = 3/1.4 + 3/4.7 + 3/7.10 +...+ 3/97.100
3A = 1 - 1/100
3A = 99/100
A=99/100:3=33/100
\(=\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{97.100}\)
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
A = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
B = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}=\frac{12}{13}\)
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{11.13}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\)
\(=1-\frac{1}{13}=\frac{12}{13}\)
#)Trả lời :
\(A=\frac{\left(140+70+42+28+20+15\right)}{420}\)
\(A=\frac{315}{420}=\frac{\left(315:105\right)}{\left(420:105\right)}=\frac{3}{4}\)
Vậy : \(A=\frac{3}{4}\)
#~Will~be~Pens~#
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{9700}=\frac{0,33x}{2009}\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{1}{1}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{100}{100}-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{2009}\)
\(\Rightarrow2009.99=100.0,33x\)
\(\Rightarrow2009.99=33x\)
\(\Rightarrow2009.99:33=x\)
\(\Rightarrow2009.3=x\)
\(\Rightarrow6027=x\)
Vậy \(x=6027\)(MK KO CHẮC NÓ ĐÚNG NHÉ )
A=1/15+1/21+1/28+....+1/190
1/2A=1/30+1/42+1/56+.....+1/380
1/2A=1/5.6+1/6.7+1/7.8+....+1/19.20
1/2A=1/5-1/6+1/6-1/7+1/7-1/8+......+1/19-1/20
1/2A=1/5-1/20
1/2A=3/20
A=3/20:1/2
A=3/10
a) \(\frac{4}{11}-\frac{7}{15}+\frac{7}{11}-\frac{5}{15}\)
\(=\left(\frac{4}{11}+\frac{7}{11}\right)-\left(\frac{7}{15}+\frac{5}{15}\right)\)
\(=1-\frac{4}{5}\)
\(=\frac{1}{5}\)
b) \(\frac{7}{3}-\frac{4}{9}-\frac{1}{3}-\frac{5}{9}\)
\(=\left(\frac{7}{3}-\frac{1}{3}\right)-\left(\frac{4}{9}+\frac{5}{9}\right)\)
\(=2-1\)
\(=1\)
c) \(\frac{1}{4}+\frac{7}{33}-\frac{5}{3}\)
\(=\frac{-1}{4}+\frac{-16}{11}\)
\(=\frac{-75}{44}\)
d) \(\frac{-3}{4}\times\frac{8}{11}-\frac{3}{11}\times\frac{1}{2}\)
\(=\frac{-6}{11}-\frac{3}{22}\)
\(=\frac{15}{22}\)
e) \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{3}-\frac{1}{15}\)
\(=\frac{4}{15}\)
\(A=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\)
\(A=\frac{3}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\right)\)
\(A=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)
\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)