Tìm x biết 1/1.3+1/3.5+1/5.7+...+1/x.(x+2)=1005/2011
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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1005}{2011}\)
\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{1005}{2011}\)
\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\frac{1}{3}-\frac{1}{x+2}=\frac{2010}{2011}\)
\(\frac{1}{x+2}=\frac{1}{3}-\frac{2010}{2011}\)
\(\frac{1}{x+2}=\frac{1}{2011}\)
\(\Rightarrow x+2=2011\)
\(x=2009\)
Đặt biểu thứ là A
2A=2/1.3+2/2.5+...+2/x.x+2
2A=1-1/3+1/3-1/5+.......+1/x-1/x+2
2A=1-1/x+2
\(\dfrac{1}{1.3}\)+ \(\dfrac{1}{3.5}\)+ \(\dfrac{1}{5.7}\)+....+\(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{1005}{2011}\)
1- \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+....+\(\dfrac{1}{x}\)- \(\dfrac{1}{x+1}\)= \(\dfrac{1005}{2011}\)
1- \(\dfrac{1}{x+1}\)= \(\dfrac{1005}{2011}\)
\(\dfrac{1}{x+1}\)= 1- \(\dfrac{1005}{2011}\)
\(\dfrac{1}{x+1}\)= \(\dfrac{1006}{2011}\)
=> x +1= 2011
=> x= 2011-1
=> x=2010
Bài này mk lm đại nha bn ! Cs j sai mong bn bỏ qua .
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 !!
.........................
= \(\frac{1}{2}\). ( \(\frac{2}{1.3}\) + \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) ... + \(\frac{2}{x.\left(x+2\right)}\) )
= \(\frac{1}{2}\) . ( 1 - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{7}\) + ... + \(\frac{1}{x}\)- \(\frac{1}{x+2}\) )
= ................
Bạn tự làm tiếp nhé ! Chúc bạn học tốt :)
Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{50}{101}\)
suy ra: \(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{50}{101}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{x+2}\right)=\frac{50}{101}\)
\(\frac{1}{1}-\frac{1}{x+2}=\frac{50}{101}:\frac{1}{2}=\frac{100}{101}\)
\(\frac{1}{x+2}=1-\frac{100}{101}=\frac{1}{101}\)
suy ra: \(x+2=101\)
suy ra: \(101-2=99\)
Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)
\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)
\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)
Suy ra: x+2=82
hay x=80
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}.\frac{x+1}{x+2}=\frac{20}{41}\)
\(\frac{x+1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\frac{x+1}{x+2}=\frac{40}{41}\)
\(x+1=40
\)
\(x=40-1\)
\(x=39\)
Đúng thì ****
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