Các câu hỏi tương tự
Tính nhanh
A=\(\frac{1\cdot5\cdot6+2\cdot10\cdot12+3\cdot15\cdot18+4\cdot20\cdot24+5\cdot25\cdot30}{1\cdot3\cdot5+2\cdot6\cdot10+3\cdot9\cdot15+4\cdot12\cdot20+5\cdot15\cdot25}\)
\(A=\frac{1.5\cdot6+2\cdot10\cdot12+3\cdot15\cdot18+4\cdot20\cdot24+5\cdot25\cdot30}{1\cdot3\cdot5+2\cdot6\cdot10+3\cdot9\cdot15+4\cdot12\cdot20+5\cdot15\cdot25}\)
Tính tổng:\(S=\frac{1}{1\cdot3}-\frac{1}{2\cdot4}+\frac{1}{3\cdot5}-\frac{1}{4\cdot6}+\frac{1}{5\cdot7}-\frac{1}{6\cdot8}+\frac{1}{7\cdot9}-\frac{1}{8\cdot10}\)
\(\frac{1}{10\cdot9}-\frac{1}{9\cdot8}-\frac{1}{8\cdot7}-\frac{1}{7\cdot6}-\frac{1}{6\cdot5}-\frac{1}{5\cdot4}-\frac{1}{4\cdot3}-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
tính:
\(\frac{1}{5\cdot6}-\frac{1}{6\cdot7}-\frac{1}{7\cdot8}-\frac{1}{8\cdot9}-...-\frac{1}{2004\cdot2005}\)
Tính \(A=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(B=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)
Help me:Tính:( 2-1 + 3-1 ) : ( 2-1 - 3-1 ) + ( 2-1 : 20 ) :23frac{left(frac{2}{5}right)^7cdot5^7+left(frac{9}{4}right)^3:left(frac{3}{16}right)^3}{2^7cdot5^2+512}frac{4^6cdot9^5+6^9cdot120}{^{8^4cdot3^{12}-6^{11}}}frac{4^5cdot9^4-2cdot6^9}{2^{10}cdot3^8+6^8cdot20}
Đọc tiếp
Help me:
Tính:
( 2-1 + 3-1 ) : ( 2-1 - 3-1 ) + ( 2-1 : 20 ) :23
\(\frac{\left(\frac{2}{5}\right)^7\cdot5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7\cdot5^2+512}\)
\(\frac{4^6\cdot9^5+6^9\cdot120}{^{8^4\cdot3^{12}-6^{11}}}\)
\(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
ai biết giải bài nay ko? giúp mình đi
\(1-\frac{1}{3\cdot5}-\frac{1}{5\cdot7}-\frac{1}{7\cdot9}-...-\frac{1}{53\cdot55}-\frac{1}{55\cdot57}\)
ai biết chỉ giúp mình. Đề thi học kì của mình mà minh ko làm được.
Tính:
a/\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
b/\(\frac{2\cdot2306}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+230}}\)c/\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+...+\frac{1}{44\cdot49}\right)\left(\frac{1-3-5-...-49}{89}\right)\)