Bài 2: Cộng, trừ số hữu tỉ

NT

\(\dfrac{2^{19}+27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

so sanh

a) \(5^{20}và2550^{10}\)

b)\(999^{10}và999999^5\)

c)\(\left(\dfrac{-1^{300}}{5}\right)và\left(\dfrac{-1^{500}}{3}\right)\)

LA
24 tháng 7 2017 lúc 15:53

\(\dfrac{2^{19}+27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(=\dfrac{2^{19}+\left(3^3\right)^3+5.3.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.4\right)^{10}}\)

\(=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{3^9.2^9.2^{10}+3^{10}.4^{10}}\)

\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{3^9.2^{19}+3^{10}.\left(2^2\right)^{10}}\)

\(=\dfrac{2^{18}.3^9.\left(2.5\right)}{3^9.2^{19}+3^{10}.2^{20}}\)

\(=\dfrac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+3.2\right)}\)

\(=\dfrac{7}{2\left(1+6\right)}\)

\(=\dfrac{7}{2.7}\)

\(=\dfrac{1}{2}\)

a) \(5^{20}và2550^{10}\)

\(5^{20}=\left(5^2\right)^{10}=25^{10}< 2550^{10}\)

=> \(5^{20}< 2550^{10}\)

b) \(999^{10}và999999^5\)

\(999^{10}=\left(999^2\right)^5=1998^5< 999999^5\)

=> \(999^{10}< 999999^5\)

c) \(\left(\dfrac{-1^{300}}{5}\right)và\left(\dfrac{-1^{500}}{3}\right)\)

\(\left(\dfrac{-1^{300}}{5}\right)=\dfrac{-1}{5}\)

\(\left(\dfrac{-1^{500}}{3}\right)=\dfrac{-1}{3}\)

\(\dfrac{-1}{5}=\dfrac{-3}{15}\)

\(\dfrac{-1}{3}=\dfrac{-5}{15}\)

=> \(\dfrac{-3}{15}>\dfrac{-5}{15}\)

=> \(\left(\dfrac{-1^{300}}{5}\right)>\left(\dfrac{-1^{500}}{3}\right)\)

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