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\(A=\dfrac{51.19-19}{19-17}=\dfrac{19\left(51-1\right)}{2}=\dfrac{19.50}{2}=475\)
\(B=\dfrac{51.13}{5.13+13.12}=\dfrac{51.13}{13\left(5+12\right)}=\dfrac{51.13}{13.17}=\dfrac{51}{17}=3\)
9: \(=\dfrac{47}{51}\cdot\dfrac{17}{94}-\dfrac{47}{51}\cdot\dfrac{53}{91}-\dfrac{53}{91}\cdot\dfrac{91}{53}+\dfrac{53}{91}\cdot\dfrac{47}{51}\)
\(=\dfrac{1}{6}-1=-\dfrac{5}{6}\)
10: \(=\dfrac{13}{19}\cdot\dfrac{19}{26}-\dfrac{13}{19}\cdot\dfrac{71}{43}+\dfrac{71}{43}\cdot\dfrac{13}{19}-\dfrac{71}{43}\cdot\dfrac{86}{71}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
Xét C = \(\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}\right)-\left(\dfrac{5}{11.12}+\dfrac{5}{12.13}+...+\dfrac{5}{98.99}\right)\)
Đặt A = \(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}\)
B = \(\dfrac{5}{11.12}+\dfrac{5}{12.13}+...+\dfrac{5}{98.99}\)
=> C = A - B
Ta có : A = \(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}\)
= 2 \(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}\right)\)
= \(2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
= \(2\left(1-\dfrac{1}{99}\right)=\dfrac{2.98}{99}=\dfrac{196}{99}\)
Ta có B = \(\dfrac{5}{11.12}+\dfrac{5}{12.13}+...+\dfrac{5}{98.99}\)
= \(5\left(\dfrac{1}{11.12}+\dfrac{1}{12.13}+...+\dfrac{1}{98.99}\right)\)
= \(5\left(\dfrac{1}{11}-\dfrac{1}{99}\right)=\dfrac{8.5}{99}=\dfrac{40}{99}\)
=> C = A - B = \(\dfrac{196-40}{99}=\dfrac{156}{99}=\dfrac{52}{33}\)
\(C=\dfrac{4}{1.3}+\dfrac{4}{3.5}+.....+\dfrac{4}{97.99}-\dfrac{5}{11.12}-\dfrac{5}{12.13}-.....-\dfrac{5}{98.99}\)
\(C=\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+.....+\dfrac{4}{97.99}\right)-\left(\dfrac{5}{11.12}+\dfrac{5}{12.13}+.....+\dfrac{5}{98.99}\right)\)\(C=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.....+\dfrac{1}{97}-\dfrac{1}{99}\right)-5\left(\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+.....+\dfrac{1}{98}-\dfrac{1}{99}\right)\)\(C=2\left(1-\dfrac{1}{99}\right)-5\left(\dfrac{1}{11}-\dfrac{1}{99}\right)\)
\(C=2.\dfrac{98}{99}-5.\dfrac{8}{99}\)
\(C=\dfrac{196}{99}-\dfrac{40}{99}=\dfrac{52}{33}\)
a,\(\frac{37.13-13}{24+37.12}=\frac{36.13}{39.12}=1\)
b,125.7.16.25.4=(125.16).(25.4).7=2000.100.7=1400000
c,8.9.14+6.17.12+19.4.18=2^3.3^2.2.7+2.3.17.2^2.3+19.2^2.3^2.2=2^3.3^2.(2.7+17+19)=72.50=3600
\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
\(=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\right)\)
\(=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=\dfrac{1}{3}-\dfrac{1}{13}\)
\(=\dfrac{10}{39}\)
nhưng khi tính trong máy tính được kết quả là \(\dfrac{5}{39}\) mà bạn
\(\dfrac{51.19-19}{19-17}.\dfrac{51.13}{5.13+13.12}\)
\(=\dfrac{19\left(51-1\right)}{19-7}.\dfrac{51.13}{13\left(5+12\right)}\)
\(=\dfrac{19.50}{2}.\dfrac{51.13}{13.17}=\dfrac{19.50}{2}.3\)
\(=\dfrac{19.50.3}{2}=\dfrac{2850}{2}=1425\)
Đó là 2 bài hả bạn