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\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
\(\frac{2.6.12}{3.8.15}=\frac{2.3.4}{3.4.3}=\frac{2}{3}\)
Các phân số kia rút gọn cũng vậy.
Vậy kết quả = \(\frac{2}{3}\)
ket qua la 2/3
ai đi qua mình tích lại cho 2 tích luôn nhé
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
\(=1+\frac{1}{3}+1+\frac{1}{15}+...+1+\frac{1}{399}.\)
\(=10+\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}\)
=\(10+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\right)\)
=\(10+\frac{1}{2}\left(1-\frac{1}{21}\right)=10+\frac{1}{2}.\frac{20}{21}=\frac{220}{21}\)
\(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+.....+\dfrac{14}{197.200}\)
\(A=\dfrac{14}{3}\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\dfrac{24}{200}=\dfrac{28}{25}\)
\(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)
\(B=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+.....\dfrac{7}{19.21}\)
\(B=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\dfrac{6}{21}=1\)
\(\frac{2.6.12+6.12.20+12.21.32}{3.8.15+9.16.25+18.28.40}\)
\(=\frac{12.\left(12+120+672\right)}{3.\left(120+1200+6720\right)}\)
\(=\frac{12.\left(12+120+672\right)}{30\left(12+120+672\right)}\)
\(=\frac{12}{30}\)
\(=\frac{2}{5}\)
\(\frac{2.6.12+6.12.20+12.21.32}{3.8.15+9.16.25+18.28.40}\)
=\(\frac{2.6.12+6.12.2.10+12.3.7.4.8}{3.8.15+9.2.8.5.5+9.2.28.40}\)
=\(\frac{144+144.10+144.56}{360+360.10+360.56}\)
=\(\frac{144\left(1+10+56\right)}{360\left(1+10+56\right)}\)
=\(\frac{144.67}{360.67}\)
=\(\frac{72.2.67}{72.5.67}\)
=\(\frac{2}{5}\)
\(C=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\)
\(C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\)
\(C=\dfrac{1}{3}-\dfrac{1}{21}\)
\(C=\dfrac{2}{7}\)
1/201 + 1/202 + ... + 1/400 > 1/400 x 200
1/201 + 1/202 + ... + 1/400 > 1/2
Vậy 1/201 + 1/202 + ... + 1/400 > 1/2
Đặt \(A=\frac{1}{201}+\frac{1}{202}+...+\frac{1}{399}+\frac{1}{400}\)
Vì \(\frac{1}{201}>\frac{1}{202}>...>\frac{1}{399}>\frac{1}{400}\)nên :
\(A< \left(\frac{1}{400}+\frac{1}{400}+...+\frac{1}{400}\right)\)( Có 200 số )
\(A< \frac{1}{400}\times200\)
\(A< \frac{200}{400}\)
\(A< \frac{1}{2}\)( Điều phải chứng minh )