Đấy cũng là đề thi của huyện mình đấy.

Đây là kết quả của mik

Như ta biết đa thức bậc 2 có dạng tổng quát là: \(ax^2+bx+c\) (trong SGK có đấy)

Suy ra: \(f\left(x-1\right)=a\left(x-1\right)^2+b\left(x-1\right)+c\)

Suy ra: \(f\left(x\right)-f\left(x-1\right)=ax^2+bx+c-a\left(x-1\right)^2-b\left(x-1\right)-c\)

\(=2ax-a+b\)(bn sử dụng hằng đẳng thức để tách \(\left(x-1\right)^2=x^2-2x+1\))

Ta có: \(2ax-a+b=x\)\(\Rightarrow\hept{\begin{cases}2a=1\\b-a=0\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=\frac{1}{2}\end{cases}}}\)

Vậy đa thức cần tìm là \(f\left(x\right)=\frac{1}{2}x^2+\frac{1}{2}x+c\)

Phần sau bn tụ áp dụng

1/1*3 + 1/3*5 + 1/5*7 + ... + 1/2007*2009

= 1/2(2/1*3 + 2/3*5 + 2/5*7 + ... + 2/2007*2009)

= 1/2(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2007 - 1/2009)

= 1/2( 1- 1/2009)

= 1/2 * 2008/2009

= 1009/2009

#)Giải :

Gọi A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2007.2009

      A = 1/2 . ( 1/1 - 1/3 + 1/3 - 1/5 + ... + 1/2007 - 1/2009

      A = 1/2 . ( 1/1 - 1/2009 )

      A = 1/2 . 2008/2009

      A = 1004/2009

#)Chúc bn học tốt :D

\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+...+\(\frac{1}{2007.2009}\)=(\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{2007.2009}\)) (*)

Ta có:\(\frac{2}{n.\left(n+2\right)}\)=\(\frac{\left(n+2\right)-n}{n.\left(n+2\right)}\)=\(\frac{n+2}{n.\left(n+2\right)}\)-\(\frac{n}{n.\left(n+2\right)}\)=\(\frac{1}{n}\)-\(\frac{1}{n+2}\)

Áp dụng vào (*),ta được:\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{2007.2009}\)=\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{2007}\)-\(\frac{1}{2009}\)=1-\(\frac{1}{2009}\)=\(\frac{2008}{2009}\)

Vậy:\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+...+\(\frac{1}{2007.2009}\)=\(\frac{2008}{2009}\)

Đặt \(A=\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{x\left(x+2\right)}\)(sửa đề)

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

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

\(\Rightarrow A=\frac{1}{2}-\frac{3}{2x+4}\)

Sửa đề: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2020}{2021}\) \(Đkxđ:\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2020}{2021}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{2020}{2021}\)

\(\Leftrightarrow\frac{x+2}{2021}=1\)

\(\Leftrightarrow x=2019\)

Vậy \(x=2019\)

\(\text{Ta có:}\) \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}\)

\(\Leftrightarrow2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right).x=\frac{2}{3}.2\)

\(\Leftrightarrow\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).x=\frac{4}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right).x=\frac{4}{3}\)

\(\Leftrightarrow\left(1-\frac{1}{11}\right)x=\frac{4}{3}\)

\(\Leftrightarrow\frac{10}{11}x=\frac{4}{3}\)

\(\Leftrightarrow x=\frac{4}{3}:\frac{10}{11}=\frac{22}{15}\)