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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) 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) \(=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)=\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).0=0\)
\(1.a,\frac{18}{24}:\frac{5}{2}+\frac{7}{-10}\)
\(=\frac{18}{24}:\frac{5}{2}+\frac{-7}{10}\)
\(=\frac{18}{24}\cdot\frac{2}{5}+\frac{-7}{10}\)
\(=\frac{3}{4}\cdot\frac{2}{5}+\frac{-7}{10}\)
\(=\frac{3}{2}\cdot\frac{1}{5}+\frac{-7}{10}\)
\(=\frac{3}{10}+\frac{-7}{10}=\frac{-4}{10}=\frac{-2}{5}\)
\(\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right]\)
\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right]\)
\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{201}\right]\cdot\left[\frac{3-2-1}{6}\right]\)
\(=\left[\frac{12}{199}-\frac{23}{200}+\frac{34}{301}\right]\cdot0=0\)
2. \(a,x+5=15\Leftrightarrow x=15-5=10\)
\(b,x-\frac{5}{3}=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{1}{9}+\frac{5}{3}\)
\(\Leftrightarrow x=\frac{1}{9}+\frac{15}{9}=\frac{16}{9}\)
c, Sửa lại đề một xíu :
\(\frac{x}{15}=\frac{-2}{3}+\frac{3}{5}\)
\(\Leftrightarrow\frac{x}{15}=\frac{-10}{15}+\frac{9}{15}\)
\(\Leftrightarrow\frac{x}{15}=\frac{-1}{15}\)
\(\Leftrightarrow x=-1\)
\(d,\frac{x}{9}< \frac{7}{x}< \frac{x}{6}(x\inℤ)\)
\(\frac{x\cdot x}{9\cdot x}< \frac{7\cdot9}{9\cdot x}< \frac{7\cdot6}{6\cdot x}\)
\(\Leftrightarrow\frac{x^2}{9x}< \frac{63}{9x}< \frac{42}{6x}\)
Tự làm nốt :>