f=1-(2.3/4+3.4/6+...+101.102/202)
A=2.3/4+3.4/6+...+101.102/202)
2a=2.3/2+3.4/3+...+101.102/101
2a=3+4+...+102
2a=100.105/2
a=100.105/4
a=2625
f=1-2625
f=-(2624)
Tớ có cái này đố các cậu
a)\(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)
b)\(\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
c)\(\frac{-3^3}{25}.\frac{75}{-21}.\frac{50}{35}\)
d)\(\frac{25.48-25.18}{20.5^3}\)
e)\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2003}\right)\)
f)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{200}\right)\)
Chúc các cậu may mắn!!
Tìm x
a) \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+.....+\left(x-100\right)=101\)
b) \(x+2x+3x+....+100x=1000\)
c) \(x+1+2\left(x+2\right)+3\left(x+3\right)+...+100\left(x+100\right)=101^2\)
d) \(\frac{1+x}{1}+\frac{1+x}{2}+\frac{1+x}{3}+...+\frac{1+x}{30}=0\)
e) \(\left(1+\frac{x}{1}\right)\left(2-\frac{x}{2}\right)\left(3-\frac{x}{3}\right)=0\)
BẠN NÀO BIẾT PHẦN NÀO THÌ GIÚP MIK NHÉ!
Thank!!
Tính \(H=\left[101-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{101}{109}\right]:\left[\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{545}\right]\)
B1:Tính hợp lí
a) \(1-\frac{1}{2}\left(1+2\right)-\frac{1}{3}\left(1+2+3\right)-...-\frac{1}{101}\left(1+2+...+101\right)\)
B2
Chứng minh \(1.3.5....99=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}....\frac{100}{2}\)
Giải nhanh nhé .Mình đag cần gấp .Cảm ơn!
Tính : \(K=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right).3^5+\left(\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}.3^9\right)+...+\left(\frac{1}{3^{97}}+\frac{1}{3^{98}}+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right).3^{101}\)
Bài 1
a rút gọn B=\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
b Chứng minh A=\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}>\frac{5}{8}\)
CMR:
a, \(100-\left(1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+..+\frac{99}{100}\)
b, \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+..+\frac{1}{200}\right)=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Giải nhanh giùm mình nhé!!!!!!!!!!!!!!
CMR: \(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{101}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+ \frac{1}{102}\right)=\frac{1}{52}+\frac{1}{53}+...+\frac{1}{102}\)
A=\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+....+\frac{1}{x\cdot\left(x+3\right)}\)
3A=\(3\cdot\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+.....\left(\frac{1}{x}-\frac{1}{x+3}\right)\)
3A=\(3\cdot\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
3A=\(\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{4620}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{101}{4620}\)
\(\frac{1}{x+3}=?\)
