Ban đầu 3 nhưng phân số cuối là 1 là sao?
Nếu là D = 3 + \(\dfrac{3}{1+2}\) + \(\dfrac{3}{1+2+3}\) + ... + \(\dfrac{3}{1+2+3+...+100}\) thì mk giải được
Cho biểu thức D= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\). Chứng minh rằng D<\(\dfrac{1}{2}\)
Tính giá trị biểu thức :
\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{3}{6}+\dfrac{4}{6}+\dfrac{5}{6}\right)-\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{4}{7}+\dfrac{5}{7}+\dfrac{6}{7}\right)+...+\left(100+...+\dfrac{99}{100}\right)\)
1. Tính giá trị biểu thức:
M=\(1+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
2.Chứng tỏ rằng :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
3.So sánh
a, A=\(\dfrac{11^5+1}{11^6+1}\) ; B=\(\dfrac{11^6+1}{11^7+1}\)
b, M=\(\dfrac{15^{10}-1}{15^9+1}\) ; N=\(\dfrac{15^9-1}{15^{10}+1}\)
Tính giá trị biểu thức:
P= 1+\(\dfrac{1}{2}\)(1+2)\(\dfrac{1}{3}\)(1+2+3)+\(\dfrac{1}{4}\)(1+2+3+4)+...+\(\dfrac{1}{2016}\)(1+2+3+...+2016)
tính giá trị biểu thức sau
A= -1 -2 +3 +4 -5 -6 +7 +8-9-10+11+12 -...........-2013-2014+2015+2016
B= [\(\dfrac{1}{2}\) -1 ] : [\(\dfrac{1}{3}\) -1] : ...................: [\(\dfrac{1}{99}\) -1} : [\(\dfrac{1}{100}\) -1]
CMR: 100- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)
So sánh
\(A=\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{100}{2^{201}}\)
và \(B=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{100^3}\)
CMR 100-(1+\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\))= (\(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\))
Bài 1. Tính giá trị biểu thức
125%.\((\dfrac{-1}{2})^2\):\((1\dfrac{5}{16}-1,5)+2008^0\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.3}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
Bài 2. Tìm x, biết
x:\(3\dfrac{1}{15}=1\dfrac{1}{12}\)
\(\dfrac{3}{4}.x=27\)
\(x.\dfrac{1}{4}=\dfrac{5}{8}.\dfrac{2}{3}\)
\(\left(x+\dfrac{1}{4}\right):\dfrac{3}{7}=\dfrac{1}{6}\)
\(\dfrac{3}{4}-2.|2x-\dfrac{2}{3}|=2\)
\(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{2}\)
