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\(42+\frac{93}{6}+\frac{87}{12}+\frac{79}{20}+\frac{69}{30}+\frac{57}{42}\)
\(=37+\left(1+\frac{93}{6}\right)+\left(1+\frac{87}{12}\right)+\left(\frac{79}{20}+1\right)+\left(1+\frac{69}{30}\right)+\left(1+\frac{57}{42}\right)\)
\(=37+\frac{99}{6}+\frac{99}{12}+\frac{99}{20}+\frac{99}{30}+\frac{99}{42}\)
\(=37+99\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=37+99\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=37+99\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=37+99\left(\frac{1}{2}-\frac{1}{7}\right)\)
\(=37+99.\frac{5}{14}\)
\(=37+35\frac{5}{14}\)
\(=72\frac{5}{14}\)
2 . 53 . 27 + 6 . 9 . 87 - 3 . 18 . 40
= 2. 53 . 9 . 3 + 2. 3 .9 .87 - 3 .2 .9 .40
= 54.53+54.87-54.40
=54(53+87-40)
=54. 100=5400
12 . 25 + 29 . 25 + 59 . 25
= 25 . ( 12+29+59)
=25 . 90
=2250
43 . 37 + 93 . 43 + 57 . 61 + 69 . 57
= 43.(37+93)+57(61+69)
=43.130+57.130
=130.(43+57)
=130.100
=13 000
12 . 53 + 53 . 172 - 53
=53(12+172-1)
=53.184
=9752
35 . 54 + 35 . 58 + 65 . 75 + 65 . 45
=35(54+58)+65(75+45)
=35.112+65.120
=3920+7800
=11720
39 . 8 + 70 . 3 + 31 . 8
=8(39+31)+70.3
=8.70+70.3
=70.(8+3)
=70.11
=770
A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9
= 1- 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9
= 1 - 1/9 = 8/9
Câu B, C dấu * là nhân hay công vậy?
\(\frac{-1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
= \(-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
=\(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
=\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
=\(-\left(1-\frac{1}{10}\right)=-\left(\frac{9}{10}\right)=-\frac{9}{10}\)
\(\frac{-1}{90}-\frac{-1}{72}-\frac{-1}{56}-\frac{-1}{42}-\frac{-1}{30}-\frac{-1}{20}-\frac{-1}{12}-\frac{-1}{6}-\frac{-1}{2}\)
\(=\frac{-1}{10.9}-\frac{-1}{9.8}-\frac{-1}{8.7}-\frac{-1}{7.6}-\frac{-1}{6.5}-\frac{-1}{5.4}-\frac{-1}{4.3}-\frac{-1}{3.2}-\frac{-1}{2.1}\)
1)
\(45\left(13+78\right)+9\left(87+22\right)5\)
\(=45\left(13+78\right)+45\left(87+22\right)\)
\(=45\left(13+78+87+22\right)\)
\(=45\left[\left(87+13\right)+\left(78+22\right)\right]\)
\(=45\left[100+100\right]\)
\(=45.200\)
\(=9000\)
2)
\(43+37+93.43+57.61+69.57\)
\(=43\left(37+93\right)+57.\left(61+69\right)\)
\(=43.130+57.130\)
\(=130\left(43+57\right)\)
\(=130.100\)
=13000
1/100 - 1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2 = -0,833
\(C=\frac{8}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{90}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{4}{45}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{4}{45}-\left(1-\frac{1}{9}\right)=\frac{4}{45}-\frac{8}{9}=\frac{4}{45}-\frac{40}{45}=\frac{-36}{45}=\frac{-4}{5}\)