Bai 5. Tinh nhanh
a, \(\dfrac{1}{5.8}+\dfrac{1}{8.7}+\dfrac{1}{11.14}+.......+\dfrac{1}{605.606}\)
b,\(\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{11}-1\right)\left(\dfrac{1}{12}-1\right)....\left(\dfrac{1}{2012}-1\right)\)
Bài 1 : Thực hiện phép tính ( tính hợp lý nếu có thể )
a ) \(\dfrac{1}{12}+\dfrac{3}{4}\)
b ) \(\dfrac{-4}{7}.1\dfrac{1}{2}\)
c )\(\dfrac{7}{9}+\left(\dfrac{2}{3}+\dfrac{-7}{9}\right)\)
d )\(\dfrac{2}{3}-\dfrac{1}{3}:\dfrac{3}{4}\)
e )\(\dfrac{-7}{25}.\dfrac{11}{13}+\dfrac{-7}{25}.\dfrac{2}{13}\)
g )\(2\dfrac{2}{5}.0,25-\left(\dfrac{11}{20}+75\%\right):\dfrac{13}{5}\)
Thực hiện phép tính
a) A= \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)\)\(+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{2013}\left(1+2+...+2013\right)\)
b) B=\(\dfrac{1-3}{1.3}+\dfrac{2-4}{2.4}+\dfrac{3-5}{3.5}+\dfrac{4-6}{4.6}+...+\dfrac{2011-2013}{2011.2013}+\dfrac{2012-2014}{2012.2014}-\dfrac{2013+2014}{2013.2014}\)
\(\left(\dfrac{1}{64}-\dfrac{1}{3^2}\right).\left(\dfrac{1}{64}-\dfrac{1}{4^2}\right).\left(\dfrac{1}{64}-\dfrac{1}{5^2}\right)...\left(\dfrac{1}{64}-\dfrac{1}{64^2}\right)\)
tìm x biết :
a) \(\left|x+\dfrac{1}{2}\right|\)=\(\dfrac{5}{2}\) b) \(\left|2x-\dfrac{2}{3}\right|\)+\(\dfrac{1}{3}\)=0 c) |x-2| = 2x + 1
cho G = \(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{16}\right)+\left(1+\dfrac{1}{256}\right)......\left(1+\dfrac{1}{2^{1024}}\right)\)và H = \(\dfrac{1}{2^{2047}}\)
Tính G + H
Cho ba phân số bằng nhau \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) . Chứng minh rằng : \(^{\left(\dfrac{a}{b}\right)^3}\)= \(\dfrac{a}{d}\)
Tìm giá trị của biểu thức :
\(A=-1,6:\left(1+\dfrac{2}{3}\right)\)
\(B=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
a) \(\dfrac{2}{1^2}.\dfrac{6}{2^2}.\dfrac{12}{3^2}.\dfrac{20}{4^2}.\dfrac{30}{5^2}.....\dfrac{110}{10^2}.x=-20\)
b) \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right).x+2013=\dfrac{2014}{1}+\dfrac{2015}{2}+...+\dfrac{4025}{2012}+\dfrac{4026}{2013}\)
c) \(\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right).x=\dfrac{2012}{51}+\dfrac{2012}{52}+...+\dfrac{2012}{99}+\dfrac{2012}{100}\)