\(^{\frac{15^{19}\cdot5^{10}}{9^{10}\cdot25^{14}}}\)
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Ta có
\(\frac{9^{15}.25^{43}}{27^{10}.5^{85}}=\frac{\left(3^2\right)^{15}.\left(5^2\right)^{43}}{\left(3^3\right)^{10}.5^{85}}=\frac{3^{30}.5^{86}}{3^{30}.5^{85}}=5\)
nhớ tick mình nha
\(\frac{9^{15}.25^{43}}{27^{10}.5^{85}}=\frac{\left(3^2\right)^{15}.\left(5^2\right)^{43}}{\left(3^3\right)^{10}.5^{85}}=\frac{3^{30}.5^{86}}{3^{30}.5^{85}}=\frac{5\left(3^{30}.5^{85}\right)}{5^{30}.5^{85}}=5\)
\(x=\frac{2^3\cdot25^2}{10^2\cdot5^3}=\frac{2^3\cdot\left(5^2\right)^2}{\left(2\cdot5\right)^2\cdot5^3}\)
\(=\frac{2^3\cdot5^4}{2^2\cdot5^2\cdot5^3}=\frac{2^3\cdot5^4}{2^2\cdot5^5}=\frac{2}{5}\)
Vậy x = 2/5
\(\dfrac{10^2\cdot5^3}{8\cdot25^2}\)
\(=\dfrac{2^2\cdot5^2\cdot5^3}{2^3\cdot\left(5^2\right)^2}\)
\(=\dfrac{2^2\cdot5^5}{2^3\cdot5^4}\)
\(=\dfrac{5}{2}\)
Ta có
\(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)
\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)
\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)
\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)
Lại có \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)
Từ (1),(2) => B>A
bn vt lớn dữ z -.-
\(\frac{15^{19}.5^{10}}{9^{10}.5^{14}}=\frac{3^{19}.5^{19}.5^{10}}{3^{20}.5^{14}}=\frac{3^{19}.5^{29}}{3^{20}.5^{14}}=\frac{5^{15}}{3}\)
\(\frac{15^{19}.5^{10}}{9^{10}.5^{14}}=\frac{3^{19}.5^{19}.5^{10}}{3^{20}.5^{14}}=\frac{3^{19}}{3^{20}}.\frac{5^{29}}{5^{14}}\)\(=\frac{1}{3}.\frac{5^{15}}{1}=\frac{5^{15}}{3}\)
Vậy ...
Học tốt ^^