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a, 2/3 của -420 là :
-420 x 2/3 = -280
Số cần tìm là :
-280 x 5/8 = -175
Vậy số cần tìm là -175
b, 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/x ( x + 2 ) = 1005 / 2011
1/2 x ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/ ( x ( x + 2 ) = 1005 / 2011
1/2 x ( 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/ x + 2 ) = 1005 / 2011
1/2 x ( 1 - 1/ x + 2 ) = 1005 / 2011
1 - 1 / x + 2 = 1005 / 2011 : 1/2
1 - 1 / x + 2 = 2010 / 2011
x + 2 / x + 2 - 1 / x + 2 = 2010 / 2011
x + 2 - 1 / x + 2 = 2010 / 2011
x + 1 / x + 2 = 2010 / 2011
+> x + 1 = 2010
x = 2010 - 1
x = 2009
+> x + 2 = 2011
x = 2011 - 2
x = 2009
Vậy x = 2009
Tk nha Đúng đó !!
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{19\cdot21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{19\cdot21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(1-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\cdot\frac{20}{21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{10}{21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{x}{14}=\frac{10}{21}-\frac{2}{-7}\)
\(\frac{x}{14}=\frac{16}{21}\)
\(\Rightarrow x\cdot=21=14\cdot16\)
\(\Rightarrow x\cdot21=224\)
\(\Rightarrow x=\frac{224}{21}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{5}{11}\)
\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{5}{11}\)
\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{5}{11}\Rightarrow1-\frac{1}{x+2}=\frac{5}{11}\div\frac{1}{2}=\frac{10}{11}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{10}{11}=\frac{1}{11}\Rightarrow x+2=11\Rightarrow x=11-2=9\)
\(\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{x+\left(x+2\right)}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{x}-\frac{1}{x+2}\)
\(=1-\frac{1}{x+2}=\frac{5}{11}\)
\(\frac{1}{x+2}=1-\frac{5}{11}=\frac{6}{11}\)
=> không có kết quả
=1/2*(1-1/3+1/3-1/5+....+1/x+1/x+2)
=1/2*(1-1/x+2)
=>1/2*x+1/x+2=20/21
Đến đó đưa về giống tìm x nha
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}=\frac{1}{x}\)
\(\Rightarrow1-\frac{1}{2005}=\frac{1}{x}\)
\(\Rightarrow\frac{2004}{2005}=\frac{1}{x}\)
tới đây tự làm nhé
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow20\left(x+2\right)=41\)
\(\Leftrightarrow x-2=\frac{41}{20}\)
\(\Leftrightarrow x=\frac{41}{20}+2\)
\(\Leftrightarrow x=\frac{81}{20}\)
\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)
Ta có :
\(\dfrac{1}{2}\)(\(\dfrac{1}{1}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\))=\(\dfrac{20}{41}\)
\(\dfrac{1}{2}\)(\(\dfrac{1}{3}\)-\(\dfrac{1}{x+2}\))=\(\dfrac{20}{41}\)
\(\dfrac{1}{3}\)-\(\dfrac{1}{x+2}\)=\(\dfrac{40}{41}\)
\(\dfrac{1}{x+2}\)=\(\dfrac{1}{3}\)-\(\dfrac{40}{41}\)
Gọi \(A=\frac{1005}{2011}\)
A=1/3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)
A=1/1.3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)
A . 2=2/1.3 + 2/3.5 + 2/5.7 +......................+2/x.(x+2)
A . 2=1/1-1/3+1/3-1/5+1/5-1/7+..............+1/x-1/x+2
A . 2=1/1+(1/3-1/3)+(1/5-1/5)+..............+(1/x-1/x)-1/x+2
A . 2=1/1-1/x+2
Suy gia:1005/2011 . 2=1/1-1/x+2
2010/2011 =1/1-1/x+2
1/x+2 =1/1-2010/2011
1/x+2 =1/2011
Suy gia:x+2=2011
x =2011-2
x =2009
tìm x : \(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\frac{1}{x}-\frac{1}{999}=\frac{1}{2}.\frac{98}{99}\)
\(\frac{1}{x}-\frac{1}{9999}=\frac{49}{99}\)
\(\frac{1}{x}=\frac{49}{99}+\frac{1}{9999}\)
\(\frac{1}{x}=\frac{50}{101}\)
\(x=1:\frac{50}{101}\)
\(x=\frac{101}{50}\)
Vậy \(x=\frac{101}{50}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\cdots+\frac{1}{x\left(x+2\right)}=\frac{5}{12}\)
\(\frac12\cdot\left(\frac11-\frac13+\frac13-\frac15+\frac15-\frac17+\cdots+\frac{1}{x}-\frac{1}{x+2}_{}\right)=\frac{5}{12}\)
\(\frac12\cdot\left(\frac11-\frac{1}{x+2}\right)=\frac{5}{12}\)
\(1-\frac{1}{x+2}=\frac{5}{12}:\frac12\)
\(1-\frac{1}{x+2}=\frac{5}{12}\cdot2\)
\(1-\frac{1}{x+2}=\frac56\)
\(\frac{1}{x+2}=1-\frac56\)
\(\frac{1}{x+2}=\frac16\)
\(\Rightarrow x+2=6\Rightarrow x=4\)
dưới mẫu là phép nhân của các số lẻ, nhưng đáp án x lại ra bằng 4, gây mâu thuẫn nên bạn xem lại đề thử có chính xác không