Câu hỏi của ĐẶNG DUY KHƯƠNG

Question

1/x×{x+1)+1/(x+1)×(x+2)+1/(x+2)×(x+3)-1/x=1/2010

Trần Thanh Phương · Accepted Answer

$\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}$$\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}$$\frac{-1}{x+3}=\frac{1}{2010}$$\Rightarrow-\left(x-3\right)=2010$$\Rightarrow x=-2013$Câu trả lời: $\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}$$\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}$$\frac{-1}{x+3}=\frac{1}{2010}$$\Rightarrow-\left(x-3\right)=2010$$\Rightarrow x=-2013$

Nguyễn Phương Uyên · Answer

$\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}$$\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}$$\Rightarrow\frac{1}{x}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}$$\Rightarrow\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+3}=\frac{1}{2010}$$\Rightarrow\frac{1}{x+3}=\frac{1}{2010}$$\Rightarrow x=2007$Câu trả lời: $\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}$$\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}$$\Rightarrow\frac{1}{x}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}$$\Rightarrow\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+3}=\frac{1}{2010}$$\Rightarrow\frac{1}{x+3}=\frac{1}{2010}$$\Rightarrow x=2007$

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