Tính hợp lý:
a) \(4\frac{3}{4}+2,37+\frac{1}{8}+1,28+3,5+3\frac{1}{12}\)
b) \(\frac{\frac{1}{22}+\frac{3}{13}+\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{1}{2}}=\frac{11.13.2}{11.13.2}\)
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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: \(=\left(\dfrac{22\cdot26+3\cdot10-65}{130}\right):\left(\dfrac{4\cdot22-2\cdot26+3\cdot143}{286}\right)\)
\(=\dfrac{537}{130}\cdot\dfrac{286}{465}=\dfrac{1969}{775}\)
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.\(\frac{\frac{5}{22}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{3}{2}}\)
\(=\frac{\left(\frac{5}{22}+\frac{3}{13}-\frac{1}{2}\right).\left(2.11.13\right)}{\left(\frac{4}{13}-\frac{2}{11}+\frac{3}{2}\right).\left(2.11.13\right)}\)
\(=\frac{65+66-143}{88-52+429}\)
\(=\frac{-12}{465}=\frac{-4}{155}\)
Câu 1;
\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)
Câu 2:
\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{3}}=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{3}}=\frac{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)
Câu 1;
13 −17 −113 23 −27 −213 ·34 −316 −364 −3256 1−14 −116 −164 +58
=13 −17 −113 2(13 −17 −113 ) ·3(14 −116 −164 −1256 )4(14 −116 −164 −1256 ) +58
=12 ·34 +58 =38 +58 =1
Câu 2: