\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\frac{1}{3}-\frac{1}{x+2}=\frac{2010}{2011}\)

\(\frac{1}{x+2}=\frac{1}{3}-\frac{2010}{2011}\)

\(\frac{1}{x+2}=\frac{1}{2011}\)

\(\Rightarrow x+2=2011\)

\(x=2009\)

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

x=2009 nha ban

chuc ban hoc gioi

tk cho minh nha

\(\dfrac{1}{1.3}\)+ \(\dfrac{1}{3.5}\)+ \(\dfrac{1}{5.7}\)+....+\(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{1005}{2011}\)

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 .

ko biết

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 :)

 Các bạn ơi! giải chi tiết ra cho mình luôn nha 

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{1005}{2011}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{1005}{2011}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1006}{2011}\)

\(\Rightarrow x+2=2011\Rightarrow x=2011-2\Rightarrow x=2009\)

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 ****

Lương Hồ Khánh Duy trả lời đúng nhưng đúng cảu bài khác

Ở đây, câu hỏi ghi x+1 bn ghi x+2

Trần Quang Trường, Lương Hồ Khánh Duy đã trả lời đúng rồi , nếu câu hỏi như bạn nói thì phép tính ko như quy luật của nó