\(Sửa:\dfrac{17}{13}+\dfrac{61}{115}-\dfrac{4}{13}+\dfrac{54}{115}+\dfrac{2004}{2005}\\ =\left(\dfrac{17}{13}-\dfrac{4}{13}\right)+\left(\dfrac{61}{115}+\dfrac{54}{115}\right)+\dfrac{2004}{2005}\\ =1+1+\dfrac{2004}{2005}=\dfrac{6014}{2005}\)
Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
Tính : A = \(\dfrac{\dfrac{1}{14}-\dfrac{1}{30}-\dfrac{1}{46}}{\dfrac{2}{35}-\dfrac{2}{75}-\dfrac{2}{115}}\):\(\frac{\frac{3}{8}-\frac{15}{17}+\frac{30}{31}}{\frac{1}{6}-\frac{20}{51}+\frac{40}{93}}\)
Chứng minh \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2005^2}< \dfrac{2004}{2005}\)
\(\dfrac{x+5}{2005}\)+\(\dfrac{x+6}{2004}\)+\(\dfrac{x+7}{2003}\)=-3
Tính nhanh
\(c,\dfrac{-5}{13}-\left(\dfrac{3}{5}+\dfrac{3}{13}-\dfrac{4}{10}\right)\)
\(d,\left(\dfrac{1}{9}-\dfrac{9}{17}\right)+\dfrac{3}{6}-\left(\dfrac{12}{17}-\dfrac{1}{2}\right)+\dfrac{5}{9}\)
\(\dfrac{2022}{2023}.\left(\dfrac{9}{13}-\dfrac{7}{11}\right)+\dfrac{2022}{2023}.\left(\dfrac{17}{13}-\dfrac{4}{17}\right)\)
\(A=\dfrac{\left(17\dfrac{1}{4}-4\dfrac{3}{16}-13\dfrac{5}{6}\right).\left(\dfrac{-4}{7}\right)+6\dfrac{3}{4}}{\left(5\dfrac{2}{7}-\dfrac{16}{3}\right):\left(6\dfrac{2}{3}-4\dfrac{1}{2}\right)}\)
Tính hợp lí:
\(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)
Giải chi tiết dùm mik nha. Thankss
