\( {1 \over 2x3} + \) \( {1 \over 3x4} +\)\( {1 \over 4x5} +\)\(... + {1 \over 97x98} +\)\({1 \over 98x99} +\)\({ 1 \over 99x100}\)
Tính:1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+...+1/(98x99)+1/(99x100)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)
Chứng minh rằng:\(1-{1 \over 2}+{1 \over 3}-...-{1 \over 1990}={1 \over 996}+{1 \over 997}+...+{1 \over 1990}\)
cm gi ???????????????????????????????
\((100+ {99 {} \over 2}+{98 {} \over 3}+...+{1 {} \over 100})/({1 {} \over 2}+{1 {} \over 3}+{1 {} \over 4}+...+{1 {} \over 101})-2\)
Bạn đăng lại câu hỏi đi. Sao mà không thấy đề gì hết
Giải các phương trình sau :
a, \({8 \over x-8} + { 11\over x-11} = {9 \over x-9} +{10 \over x-10}\)
b, \({x \over x-3} - {x \over x-5} = { x \over x-4} - { x\over x-6}\)
c, \({ 4\over x^2 - 3x + 2 } - { 3 \over 2x^2 - 6x +1 } +1 =0\)
d, \({1\over x-1} + {2\over x-2} + {3 \over x-3} = {6 \over x-6}\)
e, \({2\over 2x+1} - {3 \over 2x-1} = {4\over 4x^2 -1}\)
f, \({ 2x\over x +1 } + { 18 \over x^2 +2x-3} = {2x-5 \over x+3}\)
g, \({1 \over x-1} + { 2x^2 -5 \over x^3 -1 } = { 4 \over x^2 +x+1}\)
a, 8/x-8 + 11/x-11 = 9/x-9 + 10/ x-10
b, x/x-3 - x/x-5 = x/x-4 - x/x-6
c, 4/x^2-3x+2 - 3/2x^2-6x+1 +1 = 0
d, 1/x-1 + 2/ x-2 + 3/x-3 = 6/x-6
e, 2/2x+1 - 3/2x-1 = 4/4x^2-1
f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3
g, 1/x-1 + 2x^2 -5/x^3 -1 = 4/ x^2 +x+1
Chứng minh rằng\(A = {1 \over 101}+{1\over 102} +{1\over 103}+{1\over 104}+...+{1\over 200}>{7\over 12}\)
Tính nhanh :
A = 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + ........1/98x99 + 1/99x100 .
ta có :\(\frac{1}{1\cdot2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{2\cdot3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3\cdot4}=\frac{1}{3}-\frac{1}{4}\)
......
\(\frac{1}{99\cdot100}=\frac{1}{99}-\frac{1}{100}\)
=> \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>A=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
cho a,b,c > 0 . cm:
\(x = {1\over 4a}+{1\over 4b}+{1\over 4c} >= {1\over 2a+b+c}+{1\over 2b+c+a}+{1\over 2c+b+a}\)
1.
\(A = {3 \over 5}.{2017 \over 2016} - {3 \over 5}- {1 \over 2016} + {2 \over 5}\)
\(B = ({12 \over 199}+ {23 \over 200} - {34 \over 201}).({1\over 2} - {1 \over 3}-{1 \over 6})\)
\(C= 2{1 \over 3}+{11 \over 5}: 33-{1 \over 50}.\)(-5)2
\(D = {4^5 . 9^4 - 10 . 6^9 \over 2^10 . 3^8 + 6^8 . 28}\)
E =( -1).(-1)2.(-1)3.(-1)4......(-1)2014
2.
\(A= {5\over2.4} + {5\over4.6} + {5\over6.8} + .....+ {5\over48.50}\)
\(B={1\over3.6} + {1\over6.9} + {1\over9.12} +.....+{1\over30.33}\)
\( {1 \over 210}+{1 \over 240}+{1 \over 272}+{1 \over 306}\)