\(\frac{-100}{200}\)+\(\frac{25}{75}\)+\(\frac{17}{23}\)+\(\frac{-6}{36}\)=?
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(\(\frac{121}{122}\). \(\frac{123}{125}\)) .(\(\frac{1995}{1996}\).\(\frac{17}{16}\)- \(\frac{21}{25}\):\(\frac{16}{17}\)).(\(\frac{42}{30}\).\(\frac{75}{23}\)-\(\frac{19}{23}\).\(\frac{210}{38}\))
= (121/122 . 123/125) . ( 1995/1996 . 17/16 - 21/25 : 16/ 17) . 0
= 0
\(\frac{9}{15}< \frac{9}{17}\)
\(\frac{12}{25}=\frac{36}{75}\)
\(\frac{17}{15}< \frac{5}{4}\)
\(=87+\frac{23}{59}+74+\frac{17}{71}+12+\frac{36}{59}+25+\frac{54}{71}\)
=\(\left(87+74+12+25\right)+\left(\frac{23}{59}+\frac{36}{59}\right)+\left(\frac{17}{71}+\frac{54}{71}\right)\)
=\(198+1+1=200\)
2. a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(3^3\right)^{25}=27^{25}\)
Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)
c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)
Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
Đè thừa một số \(\frac{25}{156}\),mk ko lại đề bài nhé
\(A=1-\frac{2+3}{2\cdot3}+.....+\frac{11+12}{11\cdot12}-\frac{12+13}{12\cdot13}\)
\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-...+\frac{1}{11}+\frac{1}{12}-\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{2}-\frac{1}{13}=\frac{11}{26}\)
-100/200+25/75+17/23+-6/36
= -1/2+1/3+17/23+-1/6
= (-1/2+1/3+-1/6)+17/23
= -1/3+17/23
= 28/69
cfghhg