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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 \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)
b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)
c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)
d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)
=\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
=\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
=\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot0\)
\(=0\)
Kết quả = 0 nhé, nhớ ủng hộ mh, mh đang âm diểm
~ HOK TỐT ~
\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right)\cdot0\)
\(=0\)
a) 7/13.7/15 - 5/12.21/39 + 49/91.8/15
= 7/13. 7/15 - 5/12. 7/13 + 7/13.8/15
= 7/13. ( 7/15 - 5/12 + 8/15)
= 7/13. ( 7/15 + 8/15 - 5/12)
= 7/13. ( 1 - 5/12)
= 7/13. 7/12
= 49/156
b) ( 12/199 + 23/100 - 34/201) . ( 1/2-1/3-1/6)
= ( 12/199 + 23/100 - 34/201).0
= 0
a) \(=\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{7}{13}+\frac{7}{130}.\frac{8}{15}=\frac{7}{13}\left(\frac{7}{15}+\frac{8}{15}-\frac{5}{12}\right)=\frac{7}{13}\left(1-\frac{5}{12}\right)=\frac{7}{13}.\frac{7}{12}=\frac{48}{156}\)
b) \(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).0=0\)