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Đặt biểu thức bằng A,
Ta có:
A = 2 . ( 2/8 + 2/24 + 2/48 + ...+ 2/2400 )
A = 2 . ( 2/2.4 + 2/4.6 + 2/6.8 +...+ 2/48.50 )
A = 2. (1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 +... + 1/48 - 1/50 )
A = 2. ( 1/2 - 1/50 )
A = 2 . 12/25
A = 24/25
Vậy A = 24/25
\(E=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}...\frac{9^2}{8.10}=\frac{\left(2.3.4...9\right)^2}{1.2.\left(3.4...8\right)^2.9.10}=\frac{2^2.9^2}{1.2.9.10}=\frac{18}{10}=\frac{9}{5}\)
a)\(\frac{5}{12}+1-\frac{3}{7}+\frac{2}{24}-\frac{8}{14}\)
\(=\frac{5}{12}+1-\frac{3}{7}+\frac{1}{12}-\frac{4}{7}\)
\(=\left(\frac{5}{12}+\frac{1}{12}\right)+1-\left(\frac{3}{7}+\frac{4}{7}\right)\)
\(=\frac{1}{2}+1-1=\frac{1}{2}\)
b)\(-\frac{23}{4}+\frac{1}{2}+\frac{21}{4}-2\)
\(=\left(-\frac{23}{4}+\frac{21}{4}+\frac{2}{4}\right)-2\)
\(=-2\)
mk chỉ cần nhìn sơ qua là biết có câu dễ sao bn ko tự nghĩ đi hơi dễ rồi trừ khi bn đố tôi chục câu tiếng anh vật lí văn
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}\)+\(\frac{2}{96}\)
=\(2\)x (\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}\)\(+\frac{1}{96}\))
=\(2\)x (\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+...\)\(+\frac{1}{48}-\frac{1}{96}\))
=\(2\)x (\(1-\frac{1}{96}\))
=\(2\)x \(\frac{95}{96}\)
=\(\frac{190}{96}=\frac{95}{48}\)