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\(=-12.9\cdot34.7+12.9\cdot25.3-25.3\cdot12.9+25.3\cdot34.7\)
\(=-12.9\cdot34.7-25.3\cdot34.7\)
\(=34.7\cdot\left(-38.2\right)=-1325.54\)
![](https://rs.olm.vn/images/avt/0.png?1311)
16,07*34,7-24,7*16,07-5,7
=16,07*[34,7-24,7]-5,7
=16,07*10-5,7
=160,7-5,7
=155
16,07 x 34,7 - 24,7 x 16,07 - 5,7
= 16,07 x (34,7 - 24,7) - 5,7
= 16,07 x 10 - 5,7
= 160,7 - 5,7
= 155
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\(34,7+2,4:\left(3,26-2,51\right)\\ =34,7+2.4:0,75\\ =34,7+3,2\\ =37,9\)
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367+(-30)+1672+(-337)
= [367 + (-337) ] + (-30) + 1672
= 30 + (-30) + 1672
= 0 + 1672 = 1672
367 + (-30) + 1672 + (-337)
=367 + 1672 +[ (-30) + (-337)]
=367 + 1672 + (-367)
=[367 + (-367)] + 1672
=0 + 1672
=1672
#Học tốt
&YOUTUBER&
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\(\frac{3^4.4-3^6}{3^5.5+10.3^2.4}=\frac{3^4\left(4-3^2\right)}{3^5.5+2.5.3^2.2^2}=\frac{3^4\left(-5\right)}{3^2.5\left(3^3+2^3\right)}=\frac{3^2.3^2.5.\left(-1\right)}{3^2.5.35}=\frac{3^2\left(-1\right)}{35}=\frac{-9}{35}\)
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(2^12+2^5):(2^10+2^2)
=(2^12+5):(2^10+2)
=2^17:2^12
=2^17-12
=2^5.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+....+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}=\frac{6}{5}\)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{2}{5}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\frac{12}{25}\)
\(=\frac{24}{125}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{48\cdot50}\)
\(A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
\(A=\frac{1}{10}-\frac{1}{50}=\frac{5}{50}-\frac{1}{50}=\frac{4}{50}=\frac{2}{25}\)
Vậy \(A=\frac{2}{25}\)
= \(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
= \(\frac{1}{10}-\frac{1}{50}\)= \(\frac{2}{25}\)
\(=-21,26-566.72=-587,98\)