Chứng minh rằng :
\(\frac{1.3+2}{2^2}+\frac{2.4+2}{3^2}+\frac{3.5+2}{4^2}+...+\frac{2011.2013+2}{2011^2}< 2013\)
Tính giá trị của biểu thức sau bằng cách hợp lí nhất
D = \(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+...+\frac{2011^2}{2010.2012}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4^6}\)
=4/3.9/8.16/15.25/4096(CÁI NÀY MÌNH BIẾN NÓ VỀ PHÂN SỐ NHA)
=5/512 (BẠN THỰC HIỆN PHÉP TÍNH TỪ TRÁI SANG PHẢI NHÉ)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4^6}=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{4096}=\frac{4.9.16.25}{3.8.15.4096}=\frac{3.5}{2.3.256}=\frac{5}{2.256}=\frac{5}{512}\)
# Học tốt.!
=)))
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{59^2}{58.60}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.....\frac{59^2}{58.60}\)
\(=\frac{2^2.3^2.4^2....59^2}{1.3.2.4.3.5....58.60}\)
\(=\frac{\left(2.3.4...59\right)\left(2.3.4...59\right)}{\left(2.3.4...58\right)\left(3.4.5....60\right)}\)
\(=\frac{59.2}{60}=\frac{59}{30}\)
TÍNH:\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\).
\(A=\frac{2^2}{1.3}\cdot\frac{2^2}{2.4}\cdot\frac{2^2}{3.5}\cdot\frac{2^2}{4.6}\)
\(A=\frac{4}{3}\cdot\frac{1}{2}\cdot\frac{4}{15}\cdot\frac{1}{6}\)
\(A=\frac{4.1.4.1}{3.2.15.6}\)
\(A=\frac{4}{135}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}\)
\(=\frac{2.3.4.5}{1.2.3.4}.\frac{2.3.4.5}{3.4.5.6}\)
\(=\frac{5}{1}.\frac{2}{6}\)
\(=\frac{5}{1}.\frac{1}{3}\)
\(=\frac{5}{3}\)
Tính \(\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{2005^2}{2009.2011}+\frac{2006^2}{2011.2013}\)
\(B=\frac{1-3}{1\cdot3}+\frac{2-4}{2\cdot4}+\frac{3-5}{3\cdot5}+\frac{4-6}{4\cdot6}+............+\frac{2011-2013}{2011.2013}+\frac{2012-2014}{2012\cdot2014}-\frac{2013+2014}{2013\cdot2014}\)
Tính
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.......\frac{50^2}{49.51}\)
\(\text{= 2/1 . 2/3 . 3/2 . 3/4 . 4/3 . 4/5 ....... 50/49.50/51 }\)
Dùng phương pháp khử liên tiếp ta có
\(=\frac{2}{1}-\frac{50}{51}=\frac{52}{51}\)
\(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{100^2}{99.101}\)
Tính giá trị biểu thức:
A=\(\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+...+\frac{1004^2}{2007.2009}+\frac{1005^2}{2009.2011}+\frac{1006^2}{2011.2013}\)
\(A=\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{1006^2}{2011.2013}\)
\(\Leftrightarrow4A=\frac{2^2.1^2}{2^2-1}+\frac{2^2.2^2}{4^2-1}+...+\frac{2^2.1006^2}{2012^2-1}\)
\(=1006+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\right)\)
\(=1006+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(=1006+\frac{1}{2}\left(1-\frac{1}{2013}\right)=\frac{2026084}{2013}\)
\(\Rightarrow A=\frac{506521}{2013}\)