1/3x5+1/5x7+1/7x9+.....+1/2n-1x2n+1
mn nhanh len nhe giup mik voi
1/3x5+1/5x7+1/7x9+.........+1/2n-1x2n-2
tim x biet;
[1/3x5+1/5x7+1/7x9+...1/9x11]nhan x =8/11. giup minh nhe cac ban dep trai dep gai nhe.
\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+\frac{1}{9x11}\right)xX=\frac{8}{11}\)
\(\Rightarrow\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\right)xX=\frac{16}{11}\)
\(\Rightarrow\left(\frac{5-3}{3x5}+\frac{7-5}{5x7}+\frac{9-7}{7x9}+\frac{11-9}{9x11}\right)xX=\frac{16}{11}\)
\(\Rightarrow\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)xX=\frac{16}{11}\)
\(\Rightarrow\left(\frac{1}{3}-\frac{1}{11}\right)xX=\frac{16}{11}\Rightarrow\frac{8}{33}xX=\frac{16}{11}\)
\(\Rightarrow X=\frac{16}{11}:\frac{8}{33}=\frac{16}{11}x\frac{33}{8}=6\)
a, 2/1x3 + 2/3x5 + 2/5x7 + 2/7x9 +...+ 9/913 x 215
b,1/1x3 + 1/3x5 + 1/5x7 + 1/7x9 + 1/213 x 215
[ Giúp mik với mấy bạn ơi ai nhanh mình sẽ tick nha TvT ]
( Toán tính nhanh nha )
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều
trình bày bài giải tính nhanh
1/1x3+1/3x5+1/5x7+1/7x9...........1/101x103
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.+\frac{1}{101}-\frac{1}{103}\right)\)
\(\frac{1}{2}\left(1-\frac{1}{103}\right)=\frac{1}{2}\cdot\frac{100}{103}=\frac{50}{103}\)
xong r đó
Ta có:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{103}\right)=\frac{50}{103}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{101\cdot103}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}\cdot\frac{102}{103}=\frac{102}{206}\)
\(=\frac{51}{103}\)
tính bằng cách thuận tiện nhấtA= 1/1x3+1/3x5+1/5x7+1/7x9+......+1/19x21 Nhanh giùm mình nha! Cảm ơn trước nhé
1/1x3 + 1/3x5 + 1/5x7 + 1/7x9
=(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) chia 2
=(1-1/9)chia 2
=8/9 chia 2
=4/9
1/3x5 + 1/5x7 + 1/7x9 + ... + 1/2009x2011
A = \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\)
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+ \(\dfrac{2}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+ \(\dfrac{1}{2009}\) - \(\dfrac{1}{2011}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{2011}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2008}{6033}\)
A = \(\dfrac{1004}{6033}\)
\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{2}{7\times9}+..+\dfrac{1}{2009\times2011}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\\ =\dfrac{1}{3}-\dfrac{1}{2011}\)
Đến đây bn tự tính nhé.
\(\text {#DNamNgV}\)
\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+...+\dfrac{1}{2009\times2011}\)
\(=\dfrac{1}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{2009\times2011}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\times\left[\dfrac{1}{3}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{2011}\right]\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\times\dfrac{2008}{6033}\)
\(=\dfrac{1004}{6033}\)
tính bằng cách thuận tiện nhất A= 1/1x3+1/3x5+1/5x7+1/7x9+......+1/19x21 Nhanh giùm mình nha! Mình tick cho ai nhanh nhất nhé!
Lời giải:
$2\times A=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{19\times 21}$
$2\times A=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+...+\frac{21-19}{19\times 21}$
$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{19}-\frac{1}{21}$
$=1-\frac{1}{21}=\frac{20}{21}$
$\Rightarrow A=\frac{20}{21}: 2= \frac{10}{21}$
tinh nhanh p=\(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{2015x2017}+\frac{1}{2017x2019}\)
p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)
<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019
<=> p = 1/3 - 1/2019
<=> p = 224/673
\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2019}\right)\)
\(=\frac{112}{673}\)
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{2015.2017}+\frac{2}{2017.2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2015}-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2019}\right)=\frac{1}{2}.\frac{224}{673}=\frac{112}{673}\)