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a,
\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\\ =\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)...\left(1-\frac{1}{30}\right)\left(1+\frac{1}{30}\right)\\ =\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{31}{30}\\ =\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{29}{30}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{31}{30}\\ =\frac{1\cdot2\cdot...\cdot29}{2\cdot3\cdot...\cdot30}\cdot\frac{3\cdot4\cdot...\cdot31}{2\cdot3\cdot...\cdot30}\\ =\frac{1}{30}\cdot\frac{31}{2}=\frac{31}{60}\)
b,
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+...+\frac{100-98}{98\cdot99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\\ =\frac{1}{2}\cdot\frac{4450-1}{9900}=\frac{1}{2}\cdot\frac{4449}{9900}=\frac{4449}{19800}=\frac{1483}{6600}\)
c, (Chịu :V)
d,
\(D=\frac{1}{3}\left(\frac{3}{1\cdot2\cdot3\cdot4}+\frac{3}{2\cdot3\cdot4\cdot5}+...+\frac{3}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{4-1}{1\cdot2\cdot3\cdot4}+\frac{5-2}{2\cdot3\cdot4\cdot5}+...+\frac{30-27}{27\cdot28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+...+\frac{1}{27\cdot28\cdot29}-\frac{1}{28\cdot29\cdot30}\right)\\ =\frac{1}{3}\left(\frac{1}{6}-\frac{1}{24630}\right)\\ =\frac{228}{4105}\)
Chúc bạn học tốt nha
.
e) \(\frac{1}{7}.\frac{-3}{8}+\frac{-13}{8}.\frac{1}{7}\)
\(=\frac{1}{7}.\left[\left(-\frac{3}{8}\right)+\left(-\frac{13}{8}\right)\right]\)
\(=\frac{1}{7}.\left(-2\right)\)
\(=-\frac{2}{7}.\)
Chúc bạn học tốt!
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
a, \(\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}\)
\(=\left(-\frac{5}{12}+\frac{17}{12}\right)+\left[\frac{4}{37}+\left(-\frac{41}{37}\right)\right]\)
\(=1+\left(-1\right)\)
\(=-1\)
b, \(\frac{1}{2}+\left(-\frac{3}{5}\right):\left(-1\frac{1}{2}\right)-\left|-\frac{2}{5}\right|\)
\(=\frac{1}{2}+\left(-\frac{3}{5}\right):\left(-\frac{3}{2}\right)-\frac{2}{5}\)
\(=\frac{1}{2}+\frac{2}{5}-\frac{2}{5}\)
\(=\frac{1}{2}\)
Mấy bài còn lại tương tự bn tự làm nha tính số mũ ra xong thực hiện, lấy thừa số chung mà nhân ( H mik bận đi hc thêm rồi)
C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\frac{99}{100}\)
=\(\frac{-98}{100}=\frac{-49}{50}\)
C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1)
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A
Dễ thấy 1/2.1 = 1/1 - 1/2
1/3.2 = 1/2 - 1/3
.....................
1/99.98 = 1/98 - 1/99
1/100.99 = 1/99 - 1/100
=> cộng từng vế với vế ta
a) (-0,125)3 . (-8) = (-0,125). (-0,125)2 . (-8) = [(-0,125) . (-8)] . (-0,125)2 = 1.1/64 = 1/64
b) \(\frac{27^{15}}{9^{21}}=\frac{\left(3^3\right)^{15}}{\left(3^2\right)^{21}}=\frac{3^{45}}{3^{42}}=3^3=27\)
c) \(\left(-2,5\right)^3+\frac{2496^5}{\left(-832\right)^5}-\frac{98^{17}}{98^{16}}=-\frac{125}{8}+\left(-243\right)-98=-\frac{2853}{8}\)
d) \(\left(1-\frac{1}{3}-\frac{1}{6}\right)^2\cdot\left(16^{37}:2^{145}-1963^0\right)=\left(\frac{6}{6}-\frac{2}{6}-\frac{1}{6}\right)^2\left(16^{37}:2^{145}-1963^0\right)\)
\(\left(\frac{1}{2}\right)^2\cdot7=\frac{1}{4}\cdot7=\frac{7}{4}\)
a) \(\left(-0,125\right)^3.\left(-8\right)=\left(\frac{-1}{8}\right)^3.\left(-8\right)=\left(\frac{-1}{8}\right)^3\div\left(\frac{-1}{8}\right)=\left(\frac{-1}{8}\right)^2=\frac{1}{64}\)
b) \(27^{15}\div9^{21}=\left(3^3\right)^{15}\div\left(3^2\right)^{21}=3^{45}\div3^{42}=3^3=27\)
c) \(\left(-2,5\right)^3+2496^5\div\left(-832\right)^5-98^{17}\div98^{16}=\left(\frac{-5}{2}\right)^3-3^5-98\)
\(=\left(\frac{-125}{8}\right)-243-98=\frac{-2853}{8}\)
d) \(\left(1-\frac{1}{3}-\frac{1}{6}\right).\left(16^{37}\div2^{125}-1963^0\right)=\frac{1}{2}.\left[\left(2^4\right)^{37}\div2^{125}-1\right]\)
\(=\frac{1}{2}.\left[2^{148}\div2^{125}-1\right]=\frac{1}{2}.\left[2^{23}-1\right]=\frac{2^{23}-1}{2}\)