bạn nào giải giúp mình với

nếu đúng thì mình sẽ ***

=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

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}\right)=\frac{20}{41}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{40}{41}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)

\(=1-\frac{1}{x+2}=\frac{40}{41}\)

\(\frac{1}{x+2}\)\(=1-\frac{40}{41}\)

\(\frac{1}{x+2}=\frac{1}{41}\)

=> x+2=41

          x=41-2

          x=39

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)

\(\Leftrightarrow20\left(x+2\right)=41\)

\(\Leftrightarrow x-2=\frac{41}{20}\)

\(\Leftrightarrow x=\frac{41}{20}+2\)

\(\Leftrightarrow x=\frac{81}{20}\)

\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\) 

\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\)

\(\Rightarrow x+2=41\)

\(\Rightarrow x=41-2\)

\(\Rightarrow x=39\)

Vậy x = 39

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)

\(\Rightarrow\frac{1}{2}\left[\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}\right]=\frac{16}{17}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{17}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{17}\Rightarrow x+2=17\Rightarrow x=15\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)

Ta có:

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{8}{17}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{17}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{17}\)

\(\Rightarrow x+2=17\)

\(\Rightarrow x=15\)

Vậy \(x=15\)

xin lưu ý, h24 rất hoan nghêng các bn có cách giải khác so với ng đã giải để làm phong phú kiến thức nhân loại nhưng k chào đón cách giải giống hệt bn đã giải vì tạo sự nhàm chán

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 !!

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{\left(2x-1\right)\left(2x+1\right)}\)

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

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

\(2A=1-\frac{1}{2x+1}=\frac{2x}{2x+1}\)

\(A=\frac{x}{2x+1}\) 

Mà \(A=\frac{49}{99}\) \(\Leftrightarrow\frac{x}{2x+1}=\frac{49}{99}\Leftrightarrow x=49\)

x=49

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right)\left(2x-1\right)}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}+\frac{1}{2x+1}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)

\(\Rightarrow99x=49\left(2x+1\right)\)

\(\Rightarrow99x=98x+49\)

\(\Rightarrow x=49\)

Vậy : \(x=49\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{5}{11}\)

\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{5}{11}\)

\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{5}{11}\Rightarrow1-\frac{1}{x+2}=\frac{5}{11}\div\frac{1}{2}=\frac{10}{11}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{10}{11}=\frac{1}{11}\Rightarrow x+2=11\Rightarrow x=11-2=9\)

\(\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{x+\left(x+2\right)}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{x}-\frac{1}{x+2}\)

\(=1-\frac{1}{x+2}=\frac{5}{11}\)

\(\frac{1}{x+2}=1-\frac{5}{11}=\frac{6}{11}\)

=> không có kết quả

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{8}{7}\)

\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{16}{7}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{7}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{16}{7}\)

\(\Rightarrow\frac{1}{x+2}=-\frac{9}{7}\)

\(\Rightarrow-9\left(x+2\right)=7\)

\(\Rightarrow x+2=-\frac{7}{9}\)

\(\Rightarrow x=-\frac{25}{9}\)

Vậy \(x=-\frac{25}{9}\)