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e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
B = \(\left(-1\dfrac{1}{6}\right)\) : \(\left(\dfrac{-10}{3}+\dfrac{9}{4}\right)\) - \(\left(-\dfrac{3}{8}\right)\) : \(\left(8-\dfrac{51}{8}\right)\)
B = \(\dfrac{-7}{6}\) : \(\dfrac{-13}{12}\) - \(\left(-\dfrac{3}{8}\right)\) : \(\dfrac{13}{8}\)
B = \(\dfrac{14}{13}\) - \(\dfrac{-3}{13}\)
B = \(\dfrac{17}{13}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
Ta có: \(A=\dfrac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{28}\cdot3^{18}\cdot2}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2\cdot2^{28}\cdot3^{18}}{2^{28}\cdot3^{18}\cdot\left(5-7\cdot2\right)}\)
\(=\dfrac{2^{28}\cdot3^{18}\cdot\left(5\cdot2^2-2\right)}{2^{28}\cdot3^{18}\cdot\left(5-14\right)}\)
\(=\dfrac{20-2}{-9}=\dfrac{18}{-9}=-2\)
Bài 1:
a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)
\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)
\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)
\(=-10\cdot1.7=-17\)
b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)
\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)
\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)
\(=-\dfrac{539}{60}\)
c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)
\(=10\)
a) \(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2.3\right)^5}{\left(0,2\right)^5.\left(0,2\right)}=\dfrac{\left(0,2\right)^5.3^5}{\left(0,2\right)^5.\left(0,2\right)}=\dfrac{3^5}{0,2}=\dfrac{243}{0,2}=1215\)
c) \(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^2=2:\left(\dfrac{3}{6}-\dfrac{4}{6}\right)^2=2:\left(-\dfrac{1}{6}\right)^2=2:\dfrac{1}{36}=72\)
c; \(\dfrac{\left(0,3\right)^6.\left(0,04\right)^3}{\left(0,09\right)^4.\left(0,2\right)^4}\)
= \(\dfrac{0,3^6.0,2^6}{0,3^8,0,2^4}\)
= \(\dfrac{0,04}{0,09}\)
= \(\dfrac{4}{9}\)
d; \(\dfrac{2^3+2^4+2^5+2^6}{15^2}\)
= \(\dfrac{8+16+32+64}{225}\)
= \(\dfrac{\left(8+32\right)+\left(16+64\right)}{225}\)
= \(\dfrac{40+80}{225}\)
= \(\dfrac{120}{225}\)
= \(\dfrac{8}{15}\)