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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{x.\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)
\(\Leftrightarrow2A=\frac{2}{x.\left(x+1\right)\left(x+2\right)}+\frac{2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+...+\frac{2}{\left(x+8\right)\left(x+9\right)\left(x+10\right)}\)
\(=\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+8\right)\left(x+9\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\)
\(=\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+9\right)\left(x+10\right)}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Phân thức cuối hình như mẫu sai rồi bạn
Phải là (x+9)(x+10) mới đúng chứ
![](https://rs.olm.vn/images/avt/0.png?1311)
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+.....+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+....+\frac{1}{x+9}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{x+11-x+1}{\left(x+1\right)\left(x+11\right)}=\frac{12}{\left(x+1\right)\left(x+11\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(t=x-\frac{1}{x}\), đưa về phương trình bậc hai một ẩn : \(t^2-3t+\frac{8}{9}=0\)
Đặt \(a=x-\frac{1}{x}\), ta được: a2 - 3a + 8/9 = 0
=> (3a - 8)(3a - 1) = 0 => a = 8/3 hoặc a = 1/3
+ Với a = 8/3 \(\Rightarrow x-\frac{1}{x}=\frac{8}{3}\Rightarrow\frac{x^2-1}{x}=\frac{8}{3}\Rightarrow3x^2-3=8x\Rightarrow3x^2-8x-3=0\)
=> (3x + 1)(x - 3) = 0 => x = -1/3 hoặc x = 3
+ Với a = 1/3 \(\Rightarrow x-\frac{1}{x}=\frac{1}{3}\Rightarrow\frac{x^2-1}{x}=\frac{1}{3}\Rightarrow3x^2-3=x\Rightarrow3x^2-x-3=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{37}}{6}\\x=\frac{1-\sqrt{37}}{6}\end{cases}}\)
Vậy có 4 nghiệm \(x=\left\{3;-\frac{1}{3};\frac{1+\sqrt{37}}{6};\frac{1-\sqrt{37}}{6}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(=\frac{1+1+1+1+1}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\left(x+9\right)\left(x+11\right)}\)
\(=\frac{5}{\left(x+1\right)\left(x+11\right)\left(x+3\right)\left(x+9\right)\left(x+5\right)\left(x+7\right)}\)
\(=\frac{5}{\left(x^2+11x+x+11\right)\left(x^2+9x+3x+27\right)\left(x^2+7x+5x+35\right)}\)
\(=\frac{5}{\left(x^2+12x+11\right)\left(x^2+12x+27\right)\left(x^2+12x+35\right)}\)
A=\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}+\frac{1}{x+7}-\frac{1}{x+9}+\frac{1}{x+9}-\frac{1}{x+11}\)
Rút gọn hết đi ta có \(\frac{1}{x+1}-\frac{1}{x+11}\)=\(\frac{x+11}{\left(x+1\right).\left(x+11\right)}-\frac{x+1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{x+11-x-1}{\left(x+1\right).\left(x+11\right)}\)
A=\(\frac{10}{x^2+12x+11}\)