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\(C=0,5+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{5}{7}-\frac{4}{35}\right)+\frac{1}{41}\)
\(=\frac{15+10+12+5}{30}+\frac{25-4}{35}+\frac{1}{41}\)
\(=\frac{7}{5}+\frac{3}{5}+\frac{1}{41}\)
\(=2+\frac{1}{41}=\frac{83}{41}\)
\(D=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\left(\frac{1}{90}-\frac{1}{30}-\frac{1}{6}-\frac{1}{2}\right)+\left(-\frac{1}{72}-\frac{1}{12}\right)-\frac{1}{56}-\frac{1}{42}\)
\(=\frac{1-2-15-45}{90}+\frac{-1-6}{72}-\frac{1}{56}-\frac{1}{42}\)
\(=-\frac{61}{90}-\frac{7}{72}-\frac{1}{56}-\frac{1}{42}\)
\(=\frac{-1708-245-45-60}{2520}\)
\(=-\frac{49}{60}\)
Nghịch đảo của C là \(\frac{41}{83}\), nghịch đảo của D là \(-\frac{60}{49}\)
\(\frac{41}{83}\cdot\left(-\frac{60}{49}\right)=-\frac{2460}{4067}\)
C=0.5+1/3+0.4+5/7+1/6-4/35+1/41
C=1/2+1/3+2/5+5/7+1/6-4/35+1/41
C=(1/2+1/6+1/3)+(2/5+5/7-4/35)+1/41
C=1+1-1/41
C=2-1/41
=>C=81/41
D=1/90-1/72-1/56-1/42-1/30-1/20--1/12-1/6-1/2
=> D=1/9*10-1/8*9-1/7*8-1/6*7-1/5*6-1/4*5-1/3*4-1/2*3-1/1*2
=>D=-(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
=>D=-(1-1/10)
=>D=-9/10
ai k mh mh k lại
−190−172−156−142−130−120−112−16−12−190−172−156−142−130−120−112−16−12
=−190−(12+16+112+120+130+142+156+172)=−190−(12+16+112+120+130+142+156+172)
=−190−(11.2+12.3+13.4+14.5+15.6+16.7+17.8+18.9)=−190−(11.2+12.3+13.4+14.5+15.6+16.7+17.8+18.9)
=−190−(1−12+12−13+13−14+14−15+15−16+16−17+17−18+18−19)=−190−(1−12+12−13+13−14+14−15+15−16+16−17+17−18+18−19)
=−190−(1−19)=−190−(1−19)
=−190−89=−190−89
=−910
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
\(B=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{19.20}\)
\(B=5.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{19.20}\right)\)
\(B=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(B=5.\left(\frac{1}{10}-\frac{1}{20}\right)\)
\(B=5.\frac{1}{20}=\frac{1}{4}\)
\(C=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\)
\(4C=4.\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(4C=\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\)
\(4C=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\)
\(4C=\frac{1}{5}-\frac{1}{25}=\frac{4}{25}\)
\(C=\frac{4}{25}:4=\frac{1}{25}\)
\(A=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+..........+\frac{1}{8}.\frac{1}{9}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{8.9}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.......+\frac{1}{8}-\frac{1}{9}=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
\(B=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+....+\frac{1}{110}=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{10.11}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)
\(\text{c,d cơ bản tự làm nha }\)
A=>1.1/2.3+1.1/3.4+1.1/4.5+1.1/5.6+1.11/6.7+.1/7.8+1.1/8.9
=>1/2.3+1/3.4+1/4.5+1/6.7+1/7.8+1/8.9
=>1/2-1/3-1/4-1/5-1/6-1/7-1/8-1/9
=>1/2-1/9=>9/18-2/18=>7/18
Vậy A= 7/18