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

A=\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}+\left(\dfrac{1-3-5-7-...-49}{89}\right)\)

B=\(\dfrac{212.3^5.4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^4.49^2}{\left(125.71^3+59.14^3\right)}\)

C=\(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)

D=\(\left(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\right)+\left(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\right):\dfrac{1890}{2005}+115\)

E=13+23+...+103=3025

Tính F=23+42+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

1, P = \(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}\) - \(\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{2}{2004}}\)

2, Q = ( \(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\) + \(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\) ) : \(\dfrac{1980}{3758}\) + 155

3, A = 1.3 + 2.4 + 3.5 +....+ 97.99 + 98.100

4, B = 1.2.3 + 2.3.4. +...+ 48.49.50

5, C = \(\dfrac{1}{1.2.3.4}\) + \(\dfrac{1}{2.3.4.5}\) +...+ \(\dfrac{1}{27.28.29.30}\)

6, D = 1 + \(2^2\) + \(2^4\) + \(2^6\) + .... +\(2^{200}\)

7, E = \(\dfrac{1}{3.5}\)+ \(\dfrac{5}{5.7}\) +...+ \(\dfrac{1}{97.99}\)

Bài 1: Thực hiện phép tính:

a) ( 2- \(\dfrac{3}{2}\)). ( 2- \(\dfrac{4}{3}\)). (2- \(\dfrac{5}{4}\)). ( 2- \(\dfrac{6}{5}\))

b) \(\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-2003\)

c)( \(\dfrac{2003}{2004}+\dfrac{2004}{2003}\)):\(\dfrac{8028045}{8028024}\)

d) 4+ \(\dfrac{1}{1+\dfrac{1}{1+\dfrac{2}{1+\dfrac{3}{4}}}}\)