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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) \(\frac{49}{5}\)
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 \(\frac{A}{B}=\frac{1}{2009}\)
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
1.
a) \(\frac{11}{2}-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3-\frac{11}{2}\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=-\frac{5}{2}\)
\(\left|2x+-\frac{3}{2}\right|=-\frac{2}{3}:\left(-\frac{5}{2}\right)\)
\(\left|2x+-\frac{3}{2}\right|=\frac{4}{15}\)
\(\Rightarrow\left|2x+-\frac{3}{2}\right|\in\text{{}\frac{4}{15};-\frac{4}{15}\)}
Nếu, \(2x+\left(-\frac{3}{2}\right)=\frac{4}{15}\)
\(2x=\frac{53}{30}\)
\(x=\frac{53}{60}\)
Nếu, \(2x+\left(-\frac{3}{2}\right)=-\frac{4}{15}\)
\(2x=\frac{37}{30}\)
\(x=\frac{37}{60}\)
Vậy \(x\in\text{{}\frac{53}{60};\frac{37}{60}\)}
b) \(\left|\frac{2}{7}x-\frac{1}{5}\right|-\left|-x+\frac{4}{9}\right|=0\)
\(\left|\frac{2}{7}x-\frac{1}{5}\right|=\left|-x+\frac{4}{9}\right|\)
\(\Rightarrow\left|\frac{2}{7}x-\frac{1}{5}\right|\in\text{{}-x+\frac{4}{9};-\left(x+\frac{4}{9}\right)\)}
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-x+\frac{4}{9}\)
\(x=\frac{203}{405}\)
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-\left(-x+\frac{4}{9}\right)\)
\(\frac{2}{7}x-\frac{1}{5}=x-\frac{4}{9}\)
\(\frac{2}{7}x-x=\frac{1}{5}-\frac{4}{9}\)
\(-\frac{5}{7}x=-\frac{11}{45}\)
\(x=\frac{77}{225}\)
Vậy \(x\in\text{{}\frac{203}{405};\frac{77}{225}\)}
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=1\)
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}\)
=1
Ta có:\(23\frac{1}{3}:\frac{-1}{2^3}-13\frac{1}{3}:\frac{-1}{2^2}+5.\sqrt{\frac{9}{25}}=\frac{70}{3}:\frac{-1}{8}-\frac{40}{3}:\frac{-1}{4}+5.\frac{3}{5}\)
\(=\frac{70}{3}.\left(-8\right)-\frac{40}{3}.\left(-4\right)+3\)
\(=\frac{10}{3}.\left(-4\right).\left(2.7-4\right)+3\)
\(=\frac{-40}{3}.\left(14-4\right)+3\)
\(=\frac{-40}{3}.10+3\)
\(=\frac{-400}{3}+3\)
\(=\frac{-391}{3}\)
\(23\frac{1}{3}:\frac{-1}{2^3}-13\frac{1}{3}:\frac{-1}{2^2}+5.\sqrt{\frac{9}{25}}\)
\(=-\frac{184}{3}-\frac{-52}{3}+3\)
\(=-44+3\)
\(=-41\)