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a, \(\dfrac{10}{17}\) + \(\dfrac{5}{-13}\) - \(\dfrac{11}{25}\) + \(\dfrac{7}{17}\) - \(\dfrac{8}{13}\)
= ( \(\dfrac{10}{17}\) + \(\dfrac{7}{17}\)) - ( \(\dfrac{5}{13}\) + \(\dfrac{8}{13}\)) - \(\dfrac{11}{25}\)
= \(\dfrac{17}{17}\) - \(\dfrac{13}{13}\) - \(\dfrac{11}{25}\)
= 1 - 1 - \(\dfrac{11}{25}\)
= - \(\dfrac{11}{25}\)
b, 0,3 - \(\dfrac{93}{7}\) - 70% - \(\dfrac{4}{7}\)
= 0,3 - 0,7 - ( \(\dfrac{93}{7}+\dfrac{4}{7}\))
= - 0,4 - \(\dfrac{97}{7}\)
= - \(\dfrac{2}{5}\) - \(\dfrac{97}{7}\)
= - \(\dfrac{499}{35}\)
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\(C=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\\ 2C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\\ 2C-C=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\right)\\ C=1-\dfrac{1}{2^{2020}}=\dfrac{2^{2020}-1}{2^{2020}}\)
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Giải:
C=1/2 + 1/2^2 + 1/2^3 + ... + 1/2^2020
2C=1 + 1/2 + 1/2^2 + ... +1/2^2019
2C-C=(1+1/2+1/2^2+...+1/2^2019)-(1/2+1/2^2+1/2^3+...+1/2^2020)
C=1-1/2^2020
Chúc bạn học tốt!
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a: \(=\dfrac{20\left(1-12\right)}{30\left(-1-10\right)}=\dfrac{2}{3}\)
b: \(=\dfrac{11^9\cdot3^{18}}{3^{18}\cdot11^{11}}=\dfrac{1}{121}\)
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a: \(=\dfrac{20\left(1-12\right)}{30\left(-1-10\right)}=\dfrac{20}{30}=\dfrac{2}{3}\)
b: \(=\dfrac{11^9\cdot3^{18}}{3^{10}\cdot11^{11}\cdot3^8}=\dfrac{1}{121}\)
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a) Ta có: 2x+33=-11
nên 2x=-44
hay x=-22
b) Ta có: \(\dfrac{x}{2}=\dfrac{-49}{14}\)
nên x=-7
c) Ta có: \(\dfrac{5}{6}x+\dfrac{10}{3}=\dfrac{7}{2}\)
nên \(\dfrac{5}{6}x=\dfrac{7}{2}-\dfrac{10}{3}=\dfrac{1}{6}\)
hay \(x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\)
Viết lại đề câu b đi :( Không hiểu
a) \(\frac{33}{77}=\frac{33:11}{77:11}=\frac{3}{7}\)
c) \(\frac{2\times\left(-13\right)\times9\times10}{\left(-3\right)\times4\times\left(-5\right)\times26}=\frac{1\times1\times\left(-3\right)\times\left(-2\right)}{1\times2\times1\times\left(-2\right)}=-\frac{3}{2}\)
Kiểm tra lại hộ mình coi sai sót gì không nhé