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A, 1/(1.2)+...+1/[x(x+1)]=99
=>1-1/2+...+1/x+1/(x-1)=99
=>1-1/(x-1)=99
=>1/(x-1)=-98
=>1/(x-1)=-98/1
=>1.(-98)=(x-1).1(tích chéo)
=>x-1=-98
=>x=-97
a. \(\frac{1}{1.2}+...+\frac{1}{x.\left(x+1\right)}=99\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+...+\frac{1}{x}-\frac{1}{x+1}=99\)
\(\Rightarrow1-\frac{1}{x+1}=99\)
\(\Rightarrow\frac{1}{x+1}=1-99=-98\)
\(\Rightarrow x=\frac{1}{-98}-1\)
\(\Rightarrow x=-\frac{99}{98}\)
P/s : Bạn ơi đề sai, x sai hay mk sai ạ???
Xinloi, t ghi thiếu đề
\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)
\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)
Vì \(\left|x+\frac{1}{1.3}\right|\ge0\forall x\)
\(\left|x+\frac{1}{3.5}\right|\ge0\forall x\)
................
\(\left|x+\frac{1}{97.99}\right|\ge0\forall x\)
(VT: Vế trái; VP: Vế phải)
\(\Rightarrow VT\ge0\Rightarrow VP=50x\ge0\)mà \(50>0\)
\(\Rightarrow x>0\)
\(\Rightarrow x+\frac{1}{1.3}>0\forall x\)
..............
\(x+\frac{1}{97.99}>0\forall x\)(1)
(1) \(\Leftrightarrow x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=50x\)
\(\Leftrightarrow49x+\left(\frac{1}{1.3}+...+\frac{1}{97.99}\right)=50x\)
\(\Leftrightarrow50x-49x=\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{97.99}\right)\)
\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(\Leftrightarrow x=\frac{1}{2}\cdot\frac{98}{99}=\frac{49}{99}\)
Vậy....
P/s: Làm bừa :) Ko chắc đúng nhé
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)
\(\Leftrightarrow20\left(x+2\right)=41\)
\(\Leftrightarrow x-2=\frac{41}{20}\)
\(\Leftrightarrow x=\frac{41}{20}+2\)
\(\Leftrightarrow x=\frac{81}{20}\)
\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{5}{11}\)
\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{5}{11}\)
\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{5}{11}\Rightarrow1-\frac{1}{x+2}=\frac{5}{11}\div\frac{1}{2}=\frac{10}{11}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{10}{11}=\frac{1}{11}\Rightarrow x+2=11\Rightarrow x=11-2=9\)
\(\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{x+\left(x+2\right)}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{x}-\frac{1}{x+2}\)
\(=1-\frac{1}{x+2}=\frac{5}{11}\)
\(\frac{1}{x+2}=1-\frac{5}{11}=\frac{6}{11}\)
=> 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
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 !!