Tìm x biết:
\(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\frac{1}{2014\cdot2016}\right)\)=\(\frac{x}{1008}\)
Ta có :
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{4060225}{2014.2016}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{2015.2015}{2014.2016}\)
\(=\frac{2.3.4....2015}{1.2.3....2014}.\frac{2.3.4....2015}{3.4.5....2016}\)
\(=\frac{2015}{1}.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}=\frac{2015}{1008}\)
\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\Rightarrow x=2015\)
Vậy \(x=2015\)
Ủng hộ mk nha !!! ^_^
Rút gọn:
a,\(A=\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
b,\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2014\cdot2016}\right)\)
Chứng minh rằng với mọi số tự nhiên n\(\ge\)1:
A=\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left[1+\frac{1}{n\left(n+2\right)}\right]< 2.\)
Tính tổng của B :B=\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
HD:\(\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}\left(\frac{1}{k}+\frac{1}{k+2}\right)-\frac{1}{k+1}\)
B=1/2.1.2-1/2.2.3+1/2.2.3-1/2.3.4+...+1/2n(n+1)-1/2(n+1)(n+2)
B=1/2[(1/1.2+1/2.3+...+1/n(n+1))-(1/2.3+1/3.4+...+1/(n+1)(n+2))]
Tới đây bạn tự làm tiếp nha, tương tự như bài 1/1.2+1/2.3+..+1/n(n+1) á bạn.Cái này bạn ghi ra bạn sẽ hiểu, mình viết hơi bị lủng củng.
TÍNH TỔNG:
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
tính
C=\(\left(1+\frac{1}{1\cdot3}\right)\))\(\left(1+\frac{1}{2\cdot4}\right)+\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{2}{2015\cdot2017}\right)\)
C=(1+1+1+...+1)+(1/1*3+1/2*4+1/3*5+...+1/2015*2017+1/2015*2017)
C=2015+(2/1*3+2/2*4+2/3*5+...+2/2015*2017+2/2015*2017):2
C=2015+(1-1/3+1/2-1/4+...+1/2015-1/2017+1/2015-1/2017):2
C=2015+(1+1/2-1/2016-1/2017+1/2015-1/2017)
cai nay thi ban tu tinh lay
nho k cho minh voi nhe
=(1.3+1/1.3).(2.4+1/2.4).....(2015.2017+1/2015.2017)
=(2.2/1.3).(3.3+1/2.4)....(2016.2016/2015.2017)
=(2.3.4.....2016/1.2.3...2015).(2.3.4....2016/3.4.5...2017)
=2016.(2/2017)
=Lấy máy tính bấm nhé
(^_^)
Tính giá trị của biểu thức:
\(A=\frac{1}{2}\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2017\cdot2019}\right)\)
A=\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1.3+1}{1.3}\right).\left(\frac{2.4+1}{2.4}\right).\left(\frac{3.5+1}{3.5}\right)..........\left(\frac{2017.2019+1}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.............\frac{4072324}{2017.2019}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...................\frac{2018^2}{2017.2019}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{\left(2.3.4..........2018\right).\left(2.3.4............2018\right)}{\left(1.2.3............2017\right).\left(3.4.5..........2019\right)}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2018.2}{1.2019}\right)=\frac{2018.2}{2.2019}=\frac{2018}{2019}\)
Vậy \(A=\frac{2018}{2019}\)
Chúc bn học tốt
\(A:\frac{1}{2}=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}.\frac{3.5+1}{3.5}.....\frac{2017.2019+1}{2017.2019}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}......\frac{2018^2}{2017.2019}\)
\(=\frac{2.2.3.3.4.4.....2018.2018}{1.3.2.4.3.5....2017.2019}\)
\(=\frac{2.3.4.....2018}{1.2.3.4.....2017}.\frac{2.3.4....2018}{3.4.5.....2019}\)
\(=2018.\frac{2}{2019}\)
\(=\frac{4036}{2019}\)
\(\Rightarrow A=\frac{4036}{2019}.\frac{1}{2}\)
\(A=\frac{2018}{2019}\)
tính S1
\(S_1=\frac{1}{1\cdot2\cdot3\cdot4\cdot5}\)\(+\frac{1}{2\cdot3\cdot4\cdot5\cdot6}+.................+\frac{1}{\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(2\right)}\)
\(C=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot....\cdot\left(1-\frac{2}{99\cdot100}\right)\)
