ADMICRO
Rút gọn \(\frac{{{6^{43}}{{.25}^{21}}{{.10}^{15}}}}{{{4^{33}}{{.3}^{42}}{{.125}^{21}}.7}} \) ta được:
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ZUNIA12
Lời giải:
Báo sai\(\frac{{{6^{43}}{{.25}^{21}}{{.10}^{15}}}}{{{4^{33}}{{.3}^{42}}{{.125}^{21}}.7}} = \frac{{{{\left( {2.3} \right)}^{43}}.{{\left( {{5^2}} \right)}^{21}}.{{\left( {2.5} \right)}^{15}}}}{{{{\left( {{2^2}} \right)}^{33}}{{.3}^{42}}.{{\left( {{5^3}} \right)}^{21}}.7}} = \frac{{{2^{43}}{{.3}^{43}}{{.5}^{42}}{{.2}^{21}}{{.5}^{21}}}}{{{2^{66}}{{.3}^{42}}{{.5}^{63}}.7}} = \frac{{{2^{64}}{{.3}^{43}}{{.5}^{63}}}}{{{2^{66}}{{.3}^{42}}{{.5}^{63}}.7}} = \frac{{{2^2}.3.1}}{{1.1.1.7}} = \frac{{12}}{7}\)
ZUNIA9
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