\(A=\dfrac{1+\left(1+2\right)+\left(1+2+3\right)+.........+\left(1+2+3+......+98\right)}{1.2+2.3+3.4+.............+98.99}\) \(A=\dfrac{1+3+6+................+4851}{2+6+12+..........+9702}\)
\(A=\dfrac{1+3+6+..........+4851}{1.2+2.3+2.6+........+2.4851}\)
\(A=\dfrac{1}{2}\)
Vậy\(A=\dfrac{1}{2}\)
Tính tổng
S = \(\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+....+\dfrac{2n+1}{[n\left(n+1\right)]^2}\)
Tìm x biết:
a, \(\left|3x-4\right|\le3\)
b, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)-2x=\dfrac{1}{2}\)
Câu 1 : Tính và so sánh :
C = \(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\) so sánh với \(\dfrac{2n+2}{3n}\)
- Please help me !!!
Chứng minh rằng: \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+.................+\frac{4031}{\left(2015.2016\right)^2}< 1\)
thực hiên phép tính :
a, \(\left(3^2\right)^2-\left(2^3\right)^2-\left(-5^2\right)^2\)
b, \(2^3+3.\left(-\dfrac{1}{2}\right)^0-\left(\dfrac{1}{2}\right)^2.4+\left(\left(-2\right)^2:\dfrac{1}{2}\right):8\)
c, \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)
d, \(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
tính giá trị các biểu thức
a, A = \(\left(\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}\right):\left(\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}\right)\)
b, B = \(\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+....+\dfrac{181}{9.10}\)
CÁC BẠN ƠI GIÚP MÌNH VS????
HELP ME!!!!!!!!
Tính tổng sau
1) B= 1.2+2.3+3.4+......+99.100
2) C= \(1^2+2^2+3^2+...+99^2\)
3) D= \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{n^2}\right)\)
4) E=\(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{100}}\)
Tính tổng:
\(1.2+2.3+3.4+...+n\left(n+1\right)\)
\(1.2.3+2.3.4+3.4.5+...+n\left(n+1\right)\left(n+2\right)\)
câu 1 tính
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{2015.2017}\right)\)
