\(=\dfrac{1000+127+125}{55}=\dfrac{1252}{55}\)
\(=\dfrac{1000+127+125}{55}=\dfrac{1252}{55}\)
\(\dfrac{10^3+2.5^3+5^3}{55}\)
\(\dfrac{10^3+2.5^3+5^3}{55}-\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
tính:
\(b.\dfrac{10^3+2.5^3+5^3}{55}\)
helpppp me! mk sắp phải nọp rùi huhu
\(\frac{10^3+2.5^3+5^3}{55}\)
\(\frac{10^3+2.5^3+5^3}{55}\)
\(\frac{10^3+2.5^3+5^3}{55}\)
giúp mk vs
(\(\dfrac{3}{7}\))5 * \(\dfrac{\left(2.5\right)^{4^{ }}.7^5}{15^4.20^2}\)
\(\dfrac{(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}).230.\dfrac{1}{25}+46\dfrac{3}{4}}{(1\dfrac{3}{10}+\dfrac{10}{3}):(12\dfrac{1}{3}-14\dfrac{2}{7})}\)
Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)