\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{x\times\left(x+2\right)}=\frac{101}{102}\)

\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{101}{102}\)

\(1-\frac{1}{x+2}=\frac{101}{102}\)

\(1-\frac{1}{x+2}=1-\frac{1}{102}\)

\(\frac{1}{x+2}=\frac{1}{102}\)

x+2=102

x=102-2

x=100

2/1x3 + 2/3x5 + 2/5x7 + ... + 2/Xx(X+ 2 ) = 101/102

1/1 - 1/3 + 1/3 - 1/5  + 1/5 - 1/7 + .. + 1/x - 1/x + 2 = 101/102

1 - 1/x + 2 = 101/102

1 - 1/x + 2 = 1 - 1/102

1/x + 2 = 1/102

x + 2 =102

x = 102 - 2

x = 100

Chúc bạn học tốt!

2/1x3 + 2/3x5 + 2/5x7 + ... + 2/Xx(X+ 2 ) = 101/102

1/1 - 1/3 + 1/3 - 1/5  + 1/5 - 1/7 + .. + 1/x - 1/x + 2 = 101/102

1 - 1/x + 2 = 101/102

1 - 1/x + 2 = 1 - 1/102

1/x + 2 = 1/102

x + 2 =102

x = 102 - 2

x = 100

Chúc bạn học tốt!

a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)

Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=\dfrac{1}{100}\)

\(\Rightarrow x+1=100\)

\(x=100-1\)

\(x=99\)

câu b thiếu kết quả đúng không bn?

Công thức: \(\dfrac{1}{a\times b}=\) 1/ khoảng cách giữa a và b \(\times\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)

* Bạn làm theo công thức và vẫn dụng câu b nhé.

sửa đề câu a  và câu b  nhá  , mik nghĩ đề như này :

  \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

 \(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

= \(\frac{1}{1}-\frac{1}{215}\)

\(=\frac{214}{215}\)

b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)

    \(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)

\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)

\(A\cdot2=\frac{214}{215}\)

\(A=\frac{214}{215}:2\)

\(A=\frac{107}{215}\)

@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều

trả lời hiền thương đề bài của bạn ấy là đúm gòi nha

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{x}-\frac{2}{\left(x+2\right)}=\frac{2015}{2016}\)

\(\Rightarrow2-\frac{2}{x+2}=\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{x+2}=2-\frac{2015}{2016}\)

\(\Rightarrow\frac{2}{x+2}=\frac{2017}{2016}\)

\(\Rightarrow2017.\left(x+2\right)=2.2016\)

\(\Rightarrow2017x+4034=4032\)

\(\Rightarrow2017x=-2\)

\(\Rightarrow x=-\frac{2}{2017}\)

Vậy......

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{x\cdot\left(x+2\right)}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)

\(=1-\frac{1}{x+2}=\frac{2015}{2016}\)

=>\(\frac{1}{x+2}=\frac{1}{2016}\)

=>\(x+2=2016\)

=>\(x=2014\)

Vậy.......

Ta có:

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2015}{2016}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{2015}{2016}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{2015}{2016}=\frac{1}{2016}\)

\(\Rightarrow x+2=1\div\frac{1}{2016}=2016\)

\(\Rightarrow x=2016-2=2014\)

Vậy giá trị của x là 2014.

= 2/ 1x3 +2/ 3x5 +2/ 5x7 + 2/ 7 x9+............+2/2011 x 2003 +2 / 2003 x 2005