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− 123 + 77 + − 257 + 23 − 43 = − 123 + 23 + − 43 + − 257 + 77 = − 100 + − 300 + 77 = − 400 + 77 = − 323.
210 + 46 + – 210 + – 26 = 210 + 46 + – 210 + – 26 = 210 + – 210 + 46 + − 26 = 0 + 20 = 20.
\(\left(-123\right)+77+\left(-257\right)+23-43=\left(-123+23\right)+\left(77-257\right)-43=-100-180-43=-323\)
(−123)+77+(−257)+23−43
=(−123+23)+(77−257)−43
=−100−180−43
=−323
Ta có : (-123) + 77 + (-257) + 23 - 43
= (-123 + 23) + (77 - 257) - 43
= -100 + 180 - 43
= 80 - 43
= 37
\(\left(-123\right)+77+\left(-257\right)+23-43\)
\(=\left(-123\right)+23+\left(-257\right)+77-43\)
\(=\left(-100\right)+\left(-180\right)-43\)
\(=\left(-280\right)-43=-323\)
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a/ \(=\frac{21}{23}+\frac{125}{143}-\frac{101.21}{101.23}-\frac{1001.125}{1001.143}=0\)
b/ \(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}-\frac{10}{21}+\frac{3}{20}=\frac{7}{20}-\frac{1}{5}=\frac{4}{20}\)
c/ \(\frac{C}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{C}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{21-20}{20.21}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{C}{2}=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\Rightarrow C=\frac{19}{21}\)
B=-123+77+(-257)+23-43
=-123+77-257+23-43
=-46-257+23-43
=-303+23-43
=-280-43
=-323
