Rút gọn biểu thức
\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\left(1-\dfrac{1}{1+2+3+4+5+.....+2006}\right)\)
Giúp em với ạ
TÍNH GIÁ TRỊ BIỂU THỨC:\(\left(1+\dfrac{2}{3}\right)\cdot\left(1+\dfrac{2}{4}\right)\cdot\left(1+\dfrac{2}{5}\right)\cdot...\cdot\left(1+\dfrac{2}{97}\right)\cdot\left(1+\dfrac{2}{98}\right)\)
Tính giá trị các biểu thức sau theo cách hợp lí nhất.
a) $\mathrm{A}=\left(\dfrac{2}{7} \cdot \dfrac{1}{4}-\dfrac{1}{3} \cdot \dfrac{2}{7}\right):\left(\dfrac{2}{7} \cdot \dfrac{3}{9}-\dfrac{2}{7} \cdot \dfrac{2}{5}\right)$;
b) $\mathrm{B}=\dfrac{\left(\dfrac{1}{5}-\dfrac{2}{7}\right) \cdot \dfrac{3}{4}-\dfrac{3}{4} \cdot\left(\dfrac{1}{3}-\dfrac{2}{7}\right)}{\dfrac{1}{5} \cdot \dfrac{2}{7}-\dfrac{1}{3} \cdot\left(\dfrac{2}{7}+\dfrac{3}{9}\right)+\dfrac{3}{9} \cdot \dfrac{1}{5}} .$
1/S=\(\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
2/B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2007}\right)\)
3/C=\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
Tính:
\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)
\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)
\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)
Tính giá trị biểu thức sau:
\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)\cdot...\cdot\left(\dfrac{1}{99}-1\right)\cdot\left(\dfrac{1}{100}-1\right)\)
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right).........................\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)
\(A=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right)\left(\dfrac{1}{4}-\dfrac{4}{4}\right)................\left(\dfrac{1}{99}-\dfrac{99}{99}\right)\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(A=\left(\dfrac{-1}{2}\right)\left(\dfrac{-2}{3}\right)\left(\dfrac{-3}{4}\right)...................\left(\dfrac{-98}{99}\right)\left(\dfrac{-99}{100}\right)\)
\(A=\dfrac{\left(-1\right)\left(-2\right)\left(-3\right).........................\left(-98\right)\left(-99\right)}{2.3.4....................98.99.100}\)
\(A=\dfrac{-1}{100}\)
Ta có
A = \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)(99 thừa số)
A = \(\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)
A = \(\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right).\left(-100\right)}{2.3.4....98.99.100}\)
A = \(\dfrac{\left(-1\right).\left(-1\right).\left(-1\right)....\left(-1\right)}{1.1.1...1.1.1}\) (100 số -1, 99 số 1)
A = \(\dfrac{-1}{1.1.1.1...1.1.1}\)
A = \(\dfrac{-1}{1}\)
A = -1
Vậy A = -1
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
\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right).\cdot\cdot\cdot.\left(\dfrac{1}{99}+1\right)\) làm ơn giúp mình với
A = (\(\dfrac{1}{2}\) + 1).(\(\dfrac{1}{3}\) + 1).(\(\dfrac{1}{4}\) + 1)...(\(\dfrac{1}{99}\) + 1)
A = \(\dfrac{1+2}{2}\).\(\dfrac{1+3}{3}\).\(\dfrac{1+4}{4}\)...\(\dfrac{1+99}{99}\)
A = \(\dfrac{3}{2}\).\(\dfrac{4}{3}\).\(\dfrac{5}{4}\)....\(\dfrac{100}{99}\)
A = \(\dfrac{100}{2}\) \(\times\) \(\dfrac{3.4.5...99}{3.4.5...99}\)
A = 50
Tính giá trị của biểu thức sau:
\(D=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{2019\cdot2021}\right)\)
\(D=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{2019.2021+1}{2019.2021}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2020.2020}{2019.2021}=\left(\dfrac{2}{1}.\dfrac{3}{2}...\dfrac{2020}{2019}\right).\left(\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}\right)=2020.\dfrac{2}{2021}=\dfrac{4040}{2021}\)
1, so sánh A;B biết: A=\(\left(\dfrac{\left(3\cdot\dfrac{2}{15}+\dfrac{1}{5}\right):2\cdot\dfrac{1}{2}}{\left(5\cdot\dfrac{3}{7}-2\cdot\dfrac{1}{4}\right):\dfrac{443}{56}}\right);B=\dfrac{1,2:\left(1\cdot\dfrac{1}{5}.1\cdot\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)