\(2.\left(-3\right)^2+\left(-2\right)^3.5\)
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25 tháng 8 2023
\(R=\dfrac{\sqrt{\left(-\dfrac{2}{5}\cdot\dfrac{-5}{8}\right)^3\cdot5^2}}{\sqrt[3]{\dfrac{-3^3}{4^3}\cdot\dfrac{5^2}{2^6\cdot3^2}\cdot\dfrac{5^4}{3^4}}}\)
\(=\dfrac{\sqrt{\left(\dfrac{1}{4}\right)^3\cdot5^2}}{\sqrt[3]{\dfrac{-1}{3^3}\cdot\dfrac{25^3}{16^3}}}=\dfrac{5}{8}:\dfrac{-5}{3\cdot4}=\dfrac{5}{8}\cdot\dfrac{3\cdot4}{-5}=-\dfrac{3}{2}\)
MN
26 tháng 3 2021
a) 32 - 6 . (8 - 23) + 18 = 32 - 6 . (8 - 8) + 18
= 32 - 6 . 0 + 18 = 32 + 18 = 50
b) (3 . 5 - 9)3 . (1 + 2 . 3)2 + 42
= (15 - 9)3 . (1 + 6)2 + 42
= 63 . 72 + 42 = 216 . 49 + 16 = 10 584 + 16 = 10 600
HN
6
12 tháng 3 2020
\(2\cdot\left(-3\right)^2+\left(-2\right)^3\cdot5\)
\(=2.9+\left(-8\right)\cdot5\)
\(=18-40\)
\(=-22\)
19 tháng 10 2017
\(R=\frac{\sqrt{\left(-\frac{2}{5}\right)^5.\left(-\frac{5}{8}\right)^3.5^2}}{\sqrt[3]{\left(-\frac{3}{4}\right)^3.\left(-\frac{5}{24}\right)^2.\left(-\frac{5}{3}\right)^4}}\)
\(=\frac{\sqrt{\frac{2^5}{5^5}.\frac{5^3}{8^3}.5^2}}{-\sqrt[3]{\frac{3^3}{4^3}.\frac{5^2}{24^2}.\frac{5^4}{3^4}}}\)
\(=\frac{\sqrt{\frac{1}{16}}}{-\sqrt[3]{\frac{1}{27}.5^6.\frac{1}{2^{12}}}}=\frac{\frac{1}{4}}{-\frac{1}{3}.5^2.\frac{1}{16}}=-\frac{12}{25}\)
30 tháng 5 2022
\(R=\dfrac{\sqrt{\dfrac{-2^5\cdot\left(-5\right)^3}{5^5\cdot8^3}\cdot5^2}}{\sqrt[3]{-\dfrac{3^3}{4^3}\cdot\dfrac{5^2}{24^2}\cdot\dfrac{5^4}{3^4}}}=\dfrac{\sqrt{\dfrac{2^5\cdot5^3\cdot5^2}{5^5\cdot2^9}}}{\sqrt[3]{-\dfrac{1}{3}\cdot\dfrac{5^6}{4^3\cdot2^6\cdot3^2}}}\)
\(=\dfrac{\sqrt{\dfrac{1}{2^4}}}{\sqrt[3]{\dfrac{-1}{3^3\cdot4^3\cdot2^6}\cdot5^6}}=\dfrac{1}{2^2}:\dfrac{-5^2}{3\cdot4\cdot2^2}=\dfrac{1}{4}\cdot\dfrac{4\cdot4\cdot3}{-25}=\dfrac{-12}{25}\)
24 tháng 4 2021
Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)
Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
11 tháng 2 2022
a: \(=\dfrac{-4\cdot13\cdot9\cdot5}{3\cdot4\cdot5\cdot2\cdot13}=\dfrac{3}{2}\)
b: \(=\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot5=\dfrac{5}{6}\)
56+7.(-5)+(-9).(-2)
\(\text{56+ 7. (-5)+ (-9). (-2)}\)
\(=56+\left(-35\right)+18\)
\(=39\)
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