1/3.5+1/5.7+...+1/(2x+1)(2x+3)=100/609
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Ta có :
\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+..............+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{15}{93}\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+..............+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{30}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+..............+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}=\dfrac{30}{93}\)
\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{30}{93}\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{30}{93}=\dfrac{1}{2x+3}\)
\(\Rightarrow\dfrac{1}{93}=\dfrac{1}{2x+3}\)
\(\Rightarrow2x+3=93\)
\(2x=90\)
\(\Rightarrow x=45\)
Vậy \(x=45\) là giá trị cần tìm
~ Chúc bn học tốt ~
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(\Rightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right).\left(2x+1\right)}=2.\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)
\(\Rightarrow1-\frac{1}{2x+1}=\frac{98}{99}\)
\(\Rightarrow\frac{2x}{2x+1}=\frac{98}{99}\)
=> 2x = 98
=> x = 98 : 2 = 49
mk làm câu c cho nó dễ
c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010
=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010
=1-1/x+1=2009/2010
=1/x+1=1-2009/2010
=1/x+1=1/2010
=) x+1=2010
x =2010-1
x =2009
nhân 2 vào 2 vế rồi bạn biến đổi ra( mình lười làm ắ)
tìm được x=50 ắ
\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)
=> 2x=98
=> x=49
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow1-\dfrac{1}{2x+1}=\dfrac{98}{99}\\ \Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\Leftrightarrow198x=196x+98\\ \Leftrightarrow2x=98\Leftrightarrow x=49\)
1/3.5 + 1/5.7 +......+ 1/(2x+1)(2x+3) =100/609
2/3.5 + 2/5.7 +......+ 1/(2x+1)(2x+3)=200/609
1/3 - 1/5 + 1/5 - 1/7 +.....+1/2x+1 - 1/2x+3=200/609
1/3 - 1/2x+3 = 200/609
1/2x+3 = 1/3 - 200/609
Đoạn còn lại tự làm nhá!