So sánh :
\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}\)và \(\dfrac{1}{2}\)
Tính giá trị của biểu thức:
\(\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+...+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(A=\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+...+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{1+\left(\dfrac{1}{2013}+1\right)+\left(\dfrac{2}{2012}+1\right)+\left(\dfrac{3}{2011}+1\right)+...+\left(\dfrac{2012}{2}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{\dfrac{2014}{2014}+\dfrac{204}{2013}+\dfrac{2014}{2012}+\dfrac{2014}{2011}+...+\dfrac{2014}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{2014\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}=2014\)
mình ko chắc đúng nha !
Số số hạng của tử là :
(2013-1):1+1=2013(số hạng)
\(\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+.....+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=\dfrac{\dfrac{1}{2013}+1+\dfrac{2}{2012}+1+....+\dfrac{2012}{2}+1+\dfrac{2013}{1}-2012}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=\dfrac{\dfrac{2014}{2013}+\dfrac{2014}{2012}+....+\dfrac{2014}{2}+1}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=2014\left(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\right)\)
=2014
Mình ghi thêm ở cái dâu bằng thứ 2 cuối cùng trên tử có ghi trừ 2012 là do tử có 2013 hạng tử mà mình chỉ cộng 1 cho 2012 hạng tử nên phải trừ đi 2012
Tính giá trị biểu thức:
\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{2}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)
\(H=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2011}{1}+\dfrac{2010}{2}+...+\dfrac{1}{2011}}\)
\(I=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2012}{2}+\dfrac{2012}{3}+...+\dfrac{2012}{2011}}\)
Help me!
Chữa lại đề.Bạn xem lại đề xem đúng chưa nhé!
\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)
\(D=\dfrac{1.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}{5.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}-\dfrac{2.\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}{3\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}\)
\(D=\dfrac{1}{5}-\dfrac{2}{3}\)
\(D=-\dfrac{7}{15}\)
Cái này học lâu rồi.Bạn xem lại xem mình làm đúng chưa nhé!
Tính giá trị biểu thức
B=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2011}{1}+\dfrac{2010}{2}+\dfrac{2009}{3}+...+\dfrac{1}{ }2011}\)
Nhận xét nè: ở mẫu số tại các phân số có tử số + mẫu số = 2012. Vì vậy mục tiêu là tạo ra con 2012 ở các phân số của mẫu số. E xử con tử số ở phân số mẫu số sao cho ra con 2012 là được =))
P=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2011}{1}+\dfrac{2010}{2}+\dfrac{2009}{3}+...+\dfrac{1}{2011}}\)
Đặt B= \(\dfrac{2011}{1}+\dfrac{2010}{2}+.......+\dfrac{1}{2011}\)
Cộng 1 vào ta được:
B=(\(\dfrac{2012}{1}+\dfrac{2012}{2}+.......+\dfrac{2012}{2011}\)+\(\dfrac{2012}{2012}\)) -2012
-> B= 2012 (\(\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2011}\)+\(\dfrac{1}{2012}\)) -2012+\(\dfrac{2012}{1}\)
Thay vào P ta được:
P=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2012}}{2012\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2012}\right)}\)
-> P= \(\dfrac{1}{2012}\)
có chỗ nào chưa hiểu hỏi mình nha!
Câu 3:
a)\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
b)\(\left(\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2013}\right).x=\dfrac{2012}{1}+\dfrac{2011}{2}+\dfrac{2010}{3}+.....+\dfrac{2}{2011}+\dfrac{1}{2012}\)
a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow\)\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+2^2+2^3\right)=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+4+8\right)=480\)
\(\Leftrightarrow\)\(2^x.15=480\)
\(\Rightarrow\)\(2^x=480:15\)
\(\Leftrightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5.
Tính tổng:
S=2012+\(\dfrac{2012}{1+2}\)+\(\dfrac{2012}{1+2+3}\)+...+\(\dfrac{2012}{1+2+3+...2011}\)
3\(^{2012}\)\(-3^{2011}+3^{2010}\cdot3^{2009}+3^2-3+1\) VỚI \(\dfrac{1}{4}\)
B= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{ }2011}+\dfrac{1}{3^{2012}}\) với \(\dfrac{1}{2}\)
B=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2012}}\)
=>3B=\(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}\)
=>3B-B=2B=1-\(\dfrac{1}{3^{2012}}\)
=>B=\(\dfrac{1}{2}-\dfrac{1}{2.3^{20112}}\)<1/2
vậy........
1,
a,tính:\(\dfrac{\dfrac{7}{2012}+\dfrac{7}{9}-\dfrac{1}{4}}{\dfrac{5}{9}-\dfrac{1}{2012}-\dfrac{1}{2}}\)
b,so sánh:A=\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2010};B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{17}\)
Rút gọn A=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2013}}{2012+\dfrac{2012}{2}+\dfrac{2011}{3}+...+\dfrac{1}{2013}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)}\)
\(=\frac{1}{2014}\)
Vậy \(A=\frac{1}{2014}\)
Đặt B=\(2012+\dfrac{2012}{2}+\dfrac{2011}{3}+...+\dfrac{1}{2013}\)
=>B=\(\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{1}{2013}\right)\)
=\(\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}\)
=\(2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)\)
=>A=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2013}}{2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)}=\dfrac{1}{2014}\)
Vậy ...