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a) Ta có: \(\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}=\)\(\frac{2\cdot\left(-1\right)\cdot13\cdot3\cdot3\cdot2\cdot5}{3\cdot\left(-1\right)2\cdot2\cdot\left(-1\right)\cdot5\cdot2\cdot13}\)=\(-3.\)
b)Ta có: \(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\left(8+4\right)}{12\cdot3}\)=\(\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)
\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\left(8+4\right)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=5\)
Vậy : \(\frac{15\cdot8+15\cdot4}{12\cdot3}=5\)
a, Ta có cách làm sau
\(\dfrac{15.8+15.4}{12.3}=\dfrac{15.\left(8+4\right)}{12.3}=\dfrac{15.12}{12.3}=\dfrac{3.5.12}{12.3}\)
\(=\dfrac{5}{1}=5\)
b,
\(\dfrac{2.\left(-13\right).9.10}{-3.4.\left(-5\right).26}\)=\(\dfrac{2.\left(-13\right).\left(-3\right).\left(-3\right).\left(-2\right).\left(-5\right)}{-3.2.2_{ }.\left(-5\right).\left(-2\right).\left(-13\right)}\)=\(\dfrac{-3}{2}\)
a)Ta có:
\(\frac{99}{101}<1\)
\(1<\frac{102}{97}\)
\(\Rightarrow\frac{99}{101}<\frac{102}{97}\)
a)Tu so:
15.8+15.4
=5.3.2.4+5.3.4
=12.10+12.5
=12.15
=>\(\frac{12.15}{12.3}=\frac{12}{12}.\frac{15}{3}=\frac{15}{3}=5\)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+2^{19}.3^9.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9.\left(1+5\right)}{2^{19}.3^9\left(1+2.3\right)}\)
\(=\frac{6}{7}\)
\(\frac{15.7-15.4}{10.3}\)
\(=\frac{15.\left(7-4\right)}{10.3}\)
\(=\frac{15.3}{10.3}\)
\(=\frac{15}{10}=\frac{3}{2}\)
\(\frac{199.......9}{999.......5}=\frac{199.......9:199.......9}{999.......5:199........9}=\frac{1}{5}\)
\(\frac{121212}{424242}=\frac{121212:10101}{424242:10101}=\frac{12}{42}\)
\(\frac{187187187}{221221221}=\frac{187187187:1001001}{221221221:1001001}=\frac{187}{221}\)
\(\frac{20082008}{20112011}=\frac{20082008:10001}{20112011:10001}=\frac{2008}{2011}\)
\(\frac{199...9}{99...95}\)=\(\frac{199...9:199...9}{99...95:199...9}\)\(\frac{\left(20so9\right)}{\left(20so9\right)}\)
=\(\frac{1}{5}\)
\(\frac{3600-75}{8400-175}=\frac{75.48-75}{175.48-175}=\frac{75.\left(48-1\right)}{175.\left(48-1\right)}\)
\(=\frac{75}{175}=\frac{25.3}{25.7}=\frac{3}{7}\)
\(\frac{15.8+15.4}{12.3}=\frac{15.\left(8+4\right)}{12.3}=\frac{15.12}{12.3}=\frac{5}{1}=5\)
\(\frac{15.8+15.4}{12.3}=\frac{15.\left(8+4\right)}{12.3}=\frac{5.12}{4.3}=\frac{5}{1}=5\)