Ôn tập toán 6
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22 tháng 4 2017 lúc 21:50
chứng tỏ rằng
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}< 1\)
\(a.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
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=\(\dfrac{6}{1^2+2^2}+\dfrac{10}{2^2+3^2}+\dfrac{14}{3^2+4^2}+...+\dfrac{398}{99^2.100^2}\)
tinh:
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
Chứng minh rằng
\(\dfrac{1}{2}< \dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{199.200}< 1\)
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. \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
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/ \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{x.\left(x+1\right)}=\dfrac{2016}{2017}\)
4 Tính tổng:
B=\(\dfrac{1}{2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{99.100}\)