Giải: Ta có: \(\dfrac{1}{2.3}\)=\(\dfrac{1}{2}-\dfrac{1}{3}\) ; \(\dfrac{1}{2}-\dfrac{1}{3}\)(giữ nguyên)
Vì \(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{1}{3}\) nên \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
Vậy \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
tinh:
a. \(\dfrac{3}{4}-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
b. \(\left(-2\right)^2-1\dfrac{5}{27}.\left(-\dfrac{3}{2}\right)^3\)
c. \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
TÍnh A=\(\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)
B=\(\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+...+\dfrac{5}{98.99.100.101}\)
C=\(\dfrac{6}{1^2+2^2}+\dfrac{10}{2^2+3^2}+\dfrac{14}{3^2+4^2}+...+\dfrac{398}{99^2.100^2}\)
1) Rút gọn
A =\(\dfrac{\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+.......+\dfrac{18}{2}+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.......+\dfrac{1}{19}+\dfrac{1}{20}}\)
2) Tìm x
a/ \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{x.\left(x+1\right)}=\dfrac{2016}{2017}\)
\(a.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
So sánh A với 1 biết A = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+......+\dfrac{1}{2^{100}}\)
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
Cho biểu thức :
\(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2014}}\)
Hãy so sánh A với \(\dfrac{3}{2}\)
Bài 9 : Tính nhanh nếu có thể :
a, \(\dfrac{2^3.5^2.7^2.3^7}{49.5^3.3^6.11}\) b,\(4.\left(\dfrac{-1}{2}\right)^3-2.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)+1\)
tìm x:
a,\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right).\left(x-1\right)+\dfrac{1}{10}.x=x-\dfrac{9}{10}\)
b,\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right).\left(x-2\right)+x=\dfrac{149}{99}.x-\dfrac{98}{99}\)
