\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+......+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
Tìm x biết :\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}\left[\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\right]=\dfrac{49}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Leftrightarrow2x+1=99\Leftrightarrow x=49\)
Tìm x nguyên biết:
a)\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right).\left(2x+1\right)}=\dfrac{49}{99}\)
b)\(1-3+3^2-3^3+...+\left(-3\right)^x=\dfrac{9^{1006}-1}{2}\)
a, \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\Leftrightarrow98\left(2x+1\right)=99.2x\)
\(\Leftrightarrow2x=98\Rightarrow x=49\)
b: Đặt \(A=1-3+3^2-3^3+...+\left(-3\right)^x\)
\(=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^x\)
\(\Leftrightarrow-3A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}\)
\(\Leftrightarrow-3A-A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}-...-1\)
\(\Leftrightarrow-4A=\left(-3\right)^{x+1}-1\)
\(\Leftrightarrow A=\dfrac{\left(-3\right)^{x+1}-1}{-4}=\dfrac{-\left(-3\right)^{x+1}+1}{4}\)
\(\Leftrightarrow\dfrac{-\left(-3\right)^{x+1}+1}{4}=\dfrac{3^{2012}-1}{2}\)
\(\Leftrightarrow-\left(-3\right)^{x+1}+1=2\cdot3^{2012}-2\)
\(\Leftrightarrow-\left(-3\right)^{x+1}=2\cdot3^{2012}-3\)
\(\Leftrightarrow-\left(-3\right)^{x+1}=3\left(2\cdot3^{2011}-1\right)\)
\(\Leftrightarrow-\left(-3\right)^x=2\cdot3^{2011}-1\)
=>x=2010
Tìm x: \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\)
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{49}{99}\\ \Leftrightarrow2\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\right)=2\cdot\dfrac{49}{99}\\ \Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=1-\dfrac{1}{99}\\ \Leftrightarrow\dfrac{1}{2x+1}=\dfrac{1}{99}\\ \Rightarrow2x+1=99\\ \Leftrightarrow2x=98\\ \Leftrightarrow x=49\)
\(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) +......+ \(\dfrac{1}{(2x-1)(2x+1)}\) = \(\dfrac{49}{99}\)
HeLp me
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)
=> 2x=98
=> x=49
tìm x:
a,\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\right).\left(x-1\right)+\dfrac{1}{10}.x=x-\dfrac{9}{10}\)
b,\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right).\left(x-2\right)+x=\dfrac{149}{99}.x-\dfrac{98}{99}\)
\(\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+\left|x+\dfrac{1}{5.7}\right|+.....+\left|x+\dfrac{1}{197.199}\right|=100x\)
Tính Q =\(\dfrac{1.3}{3.5}+\dfrac{2.4}{5.7}+\dfrac{3.5}{7.9}+...+\dfrac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\dfrac{1002.1004}{2005.2007}\)
\(Tìm.x:\)
\(2x-\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2019+2021}\right)=x+1\)
ét o ét .-.
quy luật lạ z:v lúc đầu nhân lúc sau cộng:v
a)tìm các cặp số nguyên x;y thỏa mãn (2x-)(x+1)=|y+1|
b)\(\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+\left|x+\dfrac{1}{5.7}\right|+...+\left|x+\dfrac{1}{97.99}\right|=50x\)
cho A=\(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)hãy so sánh A với \(\dfrac{-1}{2}\)
b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)
Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)
Thay (1) vào đề bài:
\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)
\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)
\(\Rightarrow49x+\dfrac{16}{99}=50x\)
\(\Rightarrow x=\dfrac{16}{99}\)
Vậy \(x=\dfrac{16}{99}.\)