\(\frac{19}{4}-\frac{37}{100}+\frac{1}{8}-\frac{32}{25}-\frac{5}{2}+\frac{7}{2}=\left(\frac{19}{4}-\frac{37}{100}-\frac{32}{25}\right)+\left(\frac{7}{2}-\frac{5}{2}\right)+\frac{1}{8}=\frac{31}{10}+1+\frac{1}{8}=\frac{169}{40}\)

a.\(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(2,5\right)+3\frac{1}{12}\)

\(=\)\(\left(4\frac{3}{4}+\frac{1}{8}+3\frac{1}{12}\right)-\left(0,37+1,28+2,5\right)\)

\(=7\frac{23}{24}-4,15\)

\(=7\frac{23}{24}-4\frac{3}{20}\)

\(=3\frac{97}{120}\)

b.\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)

\(=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...\frac{1}{59}-\frac{1}{61}\right)\)

\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=\frac{3}{2}.\frac{56}{305}\)

\(=\frac{84}{305}\)

c.\(\frac{\frac{5}{22}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{3}{2}}\)

\(=\frac{\left(\frac{5}{22}+\frac{3}{13}-\frac{1}{2}\right).\left(2.11.13\right)}{\left(\frac{4}{13}-\frac{2}{11}+\frac{3}{2}\right).\left(2.11.13\right)}\)

\(=\frac{65+66-143}{88-52+429}\)

\(=\frac{-12}{465}=\frac{-4}{155}\)

a: \(=\dfrac{17}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{7}{2}\)

\(=\dfrac{35}{8}+\dfrac{8}{8}-\dfrac{37}{100}-\dfrac{128}{100}\)

\(=\dfrac{43}{8}-\dfrac{165}{100}=\dfrac{149}{40}\)

b: \(=\left(\dfrac{22\cdot26+3\cdot10-65}{130}\right):\left(\dfrac{4\cdot22-2\cdot26+3\cdot143}{286}\right)\)

\(=\dfrac{537}{130}\cdot\dfrac{286}{465}=\dfrac{1969}{775}\)

\(8\frac{2}{7}-\left(1\frac{1}{6}+25\%\right)=\frac{58}{7}-\left(\frac{7}{6}+\frac{1}{4}\right)=\frac{58}{7}-\frac{17}{12}=\frac{577}{84}\)

\(4\frac{3}{4}+\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{12}\)

\(=\frac{19}{4}+\left(-\frac{83}{20}\right)+\frac{37}{12}=\frac{3}{5}+\frac{37}{12}=\frac{221}{60}\)

\(8\frac{2}{7}-\left(1\frac{1}{6}+25\%\right)=\frac{58}{7}-\left(\frac{7}{6}+\frac{1}{4}\right)=\frac{58}{7}-\frac{17}{12}=\frac{577}{84}\)

\(8\frac{2}{7}-\left(1\frac{1}{6}+25\%\right)\)

\(=\frac{58}{7}-\left(\frac{7}{6}+\frac{1}{4}\right)\)

\(=\frac{58}{7}-\frac{17}{12}\)

\(=\frac{577}{84}\)

Câu một \(=3^2.\frac{1}{3^5}.\left(3^4\right)^2.\frac{1}{3^3}=3^{10}.\frac{1}{3^8}=3^2=9\)

Câu hai \(=\left(2^2.2^5\right):\left(2^3.\frac{1}{2^4}\right)=\frac{2^7}{\frac{2^3}{2^4}}=2^8=256\)

Chờ chút nhá :D

Câu 3  \(=9-64-25^2=-680\)

Câu 4 \(=8+1-1+1=9\)

Câu 5 \(=4,75-0,37+0,125-1,28-2,5+3\frac{1}{12}=0,725+3\frac{1}{12}=3\frac{97}{120}\)

Sai thì mình xin lỗi :v, vội quá

Câu 1:\(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}=3^2.\frac{1}{3^5}.3^8.\frac{1}{3^3}=3^2:3^5.3^8:3^3=3^{2-5+8-3}=3^2=9\)

a)\(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{117}{12}-\frac{2}{12}=\frac{115}{12}\)

b)\(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-\frac{3}{2}=\frac{13}{4}-\frac{6}{4}=\frac{7}{4}\)

c)\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}=\frac{1}{2004}\)

a. \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{39}{4}-\frac{1}{6}=\frac{115}{12}\)

b. \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-1\frac{1}{2}\)

= \(\frac{13}{4}.1-\frac{3}{2}=\frac{13}{4}-\frac{3}{2}=\frac{7}{4}\)

c. \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2004}\right)\)

= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2003}{2004}=\frac{1}{2004}\)

Giải :

a) \(\left(\frac{5}{2}-\frac{1}{3}\right).\frac{9}{2}-\frac{1}{6}=\frac{13}{6}.\frac{9}{2}-\frac{1}{6}=\frac{39}{4}-\frac{1}{6}=\frac{115}{12}\)

b) \(3\frac{1}{4}.\frac{5}{7}+\frac{2}{7}.3\frac{1}{4}-1\frac{1}{2}=3\frac{1}{4}.\left(\frac{5}{7}+\frac{2}{7}\right)-\frac{3}{2}=\frac{13}{4}-\frac{6}{4}=\frac{7}{4}\)

c) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2004}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}=\frac{1}{2004}\)

A/ \(\left(10\frac{3}{4}+3\frac{4}{5}\right)-\left(5\frac{3}{4}-1\frac{1}{5}\right)\)

\(=\left(10\frac{3}{4}-5\frac{3}{4}\right)+\left(3\frac{4}{5}+1\frac{1}{5}\right)\)

\(=5+5\)

\(=10\)

chúc bạn học tốt nha