Ta có:
A = \(\dfrac{101+100+99+98+...+1}{101-100+99-98+...+3-2+1}\)
= \(\dfrac{101+\left(100+1\right)+\left(99+2\right)+...+\left(51+50\right)}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}\)
= \(\dfrac{101+101+101+...+101}{1+1+1+...+1}\) (51 số 101 và 51 số 1)
= \(\dfrac{101.51}{51}\)
= 101
Vậy A = 101
Tính giá trị của phân số sau:
\(\dfrac{101+100+99+...+3+2+1}{101-100+99-98+...+3-2+1}\)
BT1: Tính nhanh:
2) ( \(\dfrac{98}{99}+\dfrac{89}{100}+\dfrac{100}{101}\)) . ( \(\dfrac{1}{12}-\dfrac{1}{3}+\dfrac{1}{4}\))
BT2: Tính nhanh
9) \(\dfrac{98}{99}+\dfrac{89}{100}+\dfrac{100}{101}.\left(\dfrac{1}{12}-\dfrac{1}{3}+\dfrac{1}{4}\right)\)
10) \(\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right).\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{3}{10}-\dfrac{2}{3}\right)\)
Giúp mk nha!
Tính tổng dãy số:
1+2+3-4-5-6+7+8+9-10-11-12+...+97+98+99-100-101-102
Mình cần gaaapsmong mọi người giúp đỡ
Tính \(H=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...........+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.............+\dfrac{1}{100}}:\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-..............\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+........+\dfrac{1}{500}}\)
Help me!!!
Cho B = \(\dfrac{1}{100^2}+\dfrac{1}{101^2}+....+\dfrac{1}{199^2}\)
Chứng minh:
\(\dfrac{1}{100}< B< \dfrac{1}{99}\)
a) 100-99+98-97+...+4-3+2-1
b) 99-97+95-93+...+7-5+3-1
1. 1+ (-2) + 3+ (-4) + . . . +19 + (-20)
2. 1 - 2 + 3- 4 + . . . + 99 - 100
3. -1 + 3 -5 + 7 - . . . +97 - 99
4. 1+ 2 - 3+ 4 + . . . +97 + 98 - 99 - 100
Bài 3: Tính tổng sau:
A = 101+ 103 + 105 + ...... +201
B = (-1) + 2 - 3 + 4 - 5+ 6 - ...... - 99 + 100
