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\(23\frac{1}{3}:\frac{-1}{2^3}-13\frac{1}{3}:\frac{-1}{2^2}+5.\sqrt{\frac{9}{25}}\)
\(=\frac{23.3+1}{3}:\frac{-1}{2^3}-13\frac{1}{3}:\frac{-1}{2^2}+5\sqrt{\frac{9}{25}}\)
\(=\frac{69+1}{3}:\frac{-1}{2^3}-13\frac{1}{3}:\frac{-1}{2^2}+5\sqrt{\frac{9}{25}}\)
\(=\frac{70}{3}:\frac{-1}{2^3}-13\frac{1}{3}:\frac{-1}{2^2}+5\sqrt{\frac{9}{25}}\)
\(=\frac{70}{3}:\frac{-1}{2^3}-\frac{13.3+1}{3}:\frac{-1}{2^2}+5\sqrt{\frac{9}{25}}\)
\(=\frac{70}{3}:\frac{-1}{2^3}-\frac{40}{3}:\frac{-1}{2^2}+5\sqrt{\frac{9}{25}}\)
\(=\frac{70}{3}:\frac{-1}{2^3}-\frac{40}{3}:\frac{-1}{2^2}+5.\frac{3}{5}\)
\(=\frac{70}{3}:\frac{-1}{8}-\frac{40}{3}:\frac{-1}{4}+5.\frac{3}{5}\)
\(=\frac{70}{3}.\frac{8}{-1}-\frac{40}{3}:\frac{-1}{4}+5.\frac{3}{5}\)
\(=\frac{560}{-3}-\frac{40}{3}:-\frac{1}{4}+5.\frac{3}{5}\)
\(=\frac{560}{-3}-\frac{40}{3}.\frac{4}{-1}+3\)
\(=\frac{-560}{3}-\frac{-160}{3}+\frac{9}{3}\)
\(=\frac{-391}{3}\)
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Cậu định thử sức tớ làm bài này á, có vài chỗ tớ viết tắt, chỗ nào không hiểu hỏi tớ nhé!
Tớ kiên trì lắm đấy!
\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
Ta co:\(B=\frac{2008}{1}+\frac{2007}{2}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\frac{2009-1}{1}+\frac{2009-2}{2}+...+\frac{2009-2007}{2007}+\frac{2009-2008}{2008}\)
\(B=\left(\frac{2009}{1}+\frac{2009}{2}+...+\frac{2009}{2008}\right)-\left(\frac{1}{1}+\frac{2}{2}+...+\frac{2008}{2008}\right)\)
\(B=2009+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)-2008\)
\(B=1+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2008}+\frac{1}{2009}\right)\)
Vay \(\frac{A}{B}=\frac{1}{2009}\)
Bài 1
\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)
\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)
\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)
\(=\frac{9}{25}+\frac{8}{9}-1\)
\(=\frac{56}{225}\)
\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)
\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)
\(=1:\frac{4}{3}=\frac{3}{4}\)
Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v
\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)
\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)
\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)
\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)
\(=-\frac{1}{2}\)
Ta có:\(23\frac{1}{3}:\frac{-1}{2^3}-13\frac{1}{3}:\frac{-1}{2^2}+5.\sqrt{\frac{9}{25}}=\frac{70}{3}:\frac{-1}{8}-\frac{40}{3}:\frac{-1}{4}+5.\frac{3}{5}\)
\(=\frac{70}{3}.\left(-8\right)-\frac{40}{3}.\left(-4\right)+3\)
\(=\frac{10}{3}.\left(-4\right).\left(2.7-4\right)+3\)
\(=\frac{-40}{3}.\left(14-4\right)+3\)
\(=\frac{-40}{3}.10+3\)
\(=\frac{-400}{3}+3\)
\(=\frac{-391}{3}\)
\(23\frac{1}{3}:\frac{-1}{2^3}-13\frac{1}{3}:\frac{-1}{2^2}+5.\sqrt{\frac{9}{25}}\)
\(=-\frac{184}{3}-\frac{-52}{3}+3\)
\(=-44+3\)
\(=-41\)