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Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
\(a.\)
\(-\dfrac{5}{9}\cdot\dfrac{12}{35}=\dfrac{\left(-5\right)\cdot12}{9\cdot35}=\dfrac{-60}{315}=-\dfrac{4}{21}\)
\(b.\)
\(\left(-\dfrac{5}{8}\right)\cdot-\dfrac{6}{55}=\dfrac{\left(-5\right)\cdot\left(-6\right)}{8\cdot55}=\dfrac{30}{440}=\dfrac{3}{44}\)
\(c.\)
\(\left(-7\right)\cdot\dfrac{2}{5}=-\dfrac{14}{5}\)
\(d.\)
\(-\dfrac{3}{8}\cdot\left(-6\right)=\dfrac{-3\cdot\left(-6\right)}{8}=\dfrac{18}{8}=\dfrac{9}{4}\)
\(T=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{576}\right).\left(1-\frac{1}{625}\right)\)
\(T=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{575}{576}.\frac{624}{625}\)
\(T=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{23.25}{24.24}.\frac{24.26}{25.25}\)
\(T=\frac{1.2.3...23.24}{2.3.4...24.25}.\frac{3.4.5...25.26}{2.3.4...24.25}\)
\(T=\frac{1}{25}.\frac{26}{2}=\frac{1}{25}.13=\frac{13}{25}\)
`#3107.101107`
`-3^2 + {-54 \div [-2^8 + 7] * (-2)^2}`
`= -9 + [-54 \div (-256 + 7) * 4]`
`= -9 + [-54 \div (-249) * 4]`
`= -9 + (18/83 * 4)`
`= -9 + 72/83`
`= -675/83`
______
`31 * (-18) + 31 * (-81) - 31`
`= 31 * (-18 - 81 - 1)`
`= 31 * (-100)`
`= -3100`
___
`(-12) * 47 + (-12) * 52 + (-12)`
`= (-12) * (47 + 52 + 1)`
`= (-12) * 100`
`= -1200`
___
`13 * (23 + 22) - 3 * (17 + 28)`
`= 13 * 45 - 3 * 45`
`= 45 * (13 - 3)`
`= 45 * 10`
`= 450`
____
`-48 + 48 * (-78) + 48 * (-21)`
`= 48 * (-1 - 78 - 21)`
`= 48 * (-100)`
`= -4800`
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)