Ta có: A=(1-1/2)...........................

Mà các tử có hiệu bằng 0

suy ra: Phân số có tử bằng 0

suy ra: A=0

Vậy A=0

Tích cho mk nha bn

Ta co: (1-1/2)* (1-1/3)*(1-1/4)*........*(1-1/2016)

=1/2*2/3*3/4*.....*2015/2016

=1*2*3*.....*2015/2*3*4*.....*2016

=1/2016

1/2 + 1/6+1/12 + 1/20 +....+ 1/x(x+1) = 2021/2022

1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +...+ 1/x. (x+1) = 2021/2020

1 - 1/2 + 1/2 - 1/3 + 1/3- 1/4 + 1/4 - 1/5 +...+ 1/x - 1/(x+1) = 2021/2020

1 - 1/(x+1)        = 2021/2020

     1/(x+1)         = 1 - 2021/2020

     1/(x+1)         = -1/2020

     1/(x+1)         = 1/-2020

       x + 1           = - 2020

        x                 = -2020 - 1

        x                 = -2021

Giải:

1/2+1/6+1/12+1/20+...+1/x.(x+1)=2021/2022

1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=2021/2022

1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=2021/2022

1/1-1/x+1                                    =2021/2022

      1/x+1                                    =1/1-2021/2022

      1/x+1                                    =1/2022

⇒x+1=2022

       x=2022-1

       x=2021

Chúc bạn học tốt!

\(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}=1\)

\(x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=1\)

\(x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=1\)

\(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=1\)

\(x-\left(1-\frac{1}{7}\right)=1\)

\(x-\frac{6}{7}=1\)

\(x=1+\frac{6}{7}\)

\(x=\frac{7}{7}+\frac{6}{7}=\frac{13}{7}\)

Vậy \(x=\frac{13}{7}\)

\(=\frac{13}{7}\)

ĐK : \(x\ne-2.-3;-4;-5;-6\)

\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\Leftrightarrow\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\Leftrightarrow x^2+8x-20=0\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\Leftrightarrow x=2;x=-10\)( tmđkxđ )

Vậy tập nghiệm phương trình là S = { -10 ; 2 } 

ĐKXĐ \(x\notin\left\{-2;-3;...;-6\right\}\)

Phương trình tương đương với:

\(\dfrac{1}{\left(x^2+2x\right)+\left(3x+6\right)}+\dfrac{1}{\left(x^2+3x\right)+\left(4x+12\right)}+\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}+\dfrac{\left(x+4\right)-\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}+\dfrac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\dfrac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\\\Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\\ \Leftrightarrow\begin{matrix}x=2\\x=-10\end{matrix}\left(t.m\right)\)

ĐKXĐ: \(x\notin\left\{-2;-3;-4;-5;-6\right\}\)

Ta có: \(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{8\left(x+6\right)}{8\left(x+2\right)\left(x+6\right)}-\dfrac{8\left(x+2\right)}{8\left(x+2\right)\left(x+6\right)}=\dfrac{\left(x+2\right)\left(x+6\right)}{8\left(x+2\right)\left(x+6\right)}\)

Suy ra: \(8x+48-8x-16=x^2+8x+12\)

\(\Leftrightarrow x^2+8x+12-32=0\)

\(\Leftrightarrow x^2+8x-20=0\)

\(\Leftrightarrow x^2+10x-2x-20=0\)

\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)

\(\Leftrightarrow\left(x+10\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+10=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-10;2}

Từ GT ; ta có :  \(\left(x-1\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=224\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)=224\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=224\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=224\)

\(\Rightarrow\left(x-1\right).\dfrac{2}{9}=224\)

\(\Rightarrow\left(x-1\right)=1008\)

\(\Rightarrow x=1009\)

Vậy ... 

: A = 1/6+1/12+1/20+1/30+.........+1/210

A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/14.15

A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/14 - 1/15

A = 1/2 - 1/15

A = 13/30

Ta có: 1/2= 1/1- 1/2 
1/6= 1/2 - 1/3 
1/12= 1/3- 1/4 
... 
1/30= 1/5 - 1/6 
1/42= 1/6 - 1/7 
Thay vào tổng kia: 1/2+1/6+...+1/30+1/42= 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/5 - 1/6 + 1/6 - 1/7 = 1/2 - 1/7= 5/14 
Chúc bạn học tốt. Thân!

ể! Hai câu trả lời khác nhau mình nên nghe ai đây?

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