Câu hỏi của Nguyễn Nhật thành

Question

1+$\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x.\left(x+1\right)}=1\frac{2008}{2010}$

Hoàng C5 · Accepted Answer

$1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=1\frac{2008}{2010}$$\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2008}{2010}$$\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=1\frac{2008}{2010}$:2$\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}$$\Rightarrow1-\frac{1}{x+1}=\frac{2009}{2010}$$\Rightarrow1-\frac{2009}{2010}=\frac{1}{x+1}$$\Rightarrow\frac{1}{2010}=\frac{1}{x+1}$$\Rightarrow x=2009$ nha !Câu trả lời: $1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=1\frac{2008}{2010}$$\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2008}{2010}$$\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=1\frac{2008}{2010}$:2$\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}$$\Rightarrow1-\frac{1}{x+1}=\frac{2009}{2010}$$\Rightarrow1-\frac{2009}{2010}=\frac{1}{x+1}$$\Rightarrow\frac{1}{2010}=\frac{1}{x+1}$$\Rightarrow x=2009$ nha !

Mạc Thu Hà · Answer

Ta có :A=1+$\frac{2}{6}$+$\frac{2}{12}$+......+$\frac{2}{x\left(x+1\right)}$=$\frac{4018}{2010}$$\Rightarrow$A=$\frac{2}{2.3}$+$\frac{2}{3.4}$+...+$\frac{2}{x\left(x+1\right)}$=$\frac{2008}{2010}$$\Rightarrow$A=2($\frac{1}{2.3}$+$\frac{1}{3.4}$+...+$\frac{1}{x\left(x+1\right)}$)=$\frac{2008}{2010}$$\Rightarrow$A=2($\frac{1}{2}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+...+$\frac{1}{x}$-$\frac{1}{x+1}$)=$\frac{2008}{2010}$$\Rightarrow$A=2($\frac{1}{2}$-$\frac{1}{x+1}$)=$\frac{2008}{2010}$$\Rightarrow$A=$\frac{1}{2}$-$\frac{1}{x+1}$=$\frac{502}{1005}$$\Rightarrow$$\frac{1}{x+1}$=$\frac{1}{2010}$$\Rightarrow$x+1=2010$\Rightarrow$x=2009Câu trả lời: Ta có :A=1+$\frac{2}{6}$+$\frac{2}{12}$+......+$\frac{2}{x\left(x+1\right)}$=$\frac{4018}{2010}$$\Rightarrow$A=$\frac{2}{2.3}$+$\frac{2}{3.4}$+...+$\frac{2}{x\left(x+1\right)}$=$\frac{2008}{2010}$$\Rightarrow$A=2($\frac{1}{2.3}$+$\frac{1}{3.4}$+...+$\frac{1}{x\left(x+1\right)}$)=$\frac{2008}{2010}$$\Rightarrow$A=2($\frac{1}{2}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+...+$\frac{1}{x}$-$\frac{1}{x+1}$)=$\frac{2008}{2010}$$\Rightarrow$A=2($\frac{1}{2}$-$\frac{1}{x+1}$)=$\frac{2008}{2010}$$\Rightarrow$A=$\frac{1}{2}$-$\frac{1}{x+1}$=$\frac{502}{1005}$$\Rightarrow$$\frac{1}{x+1}$=$\frac{1}{2010}$$\Rightarrow$x+1=2010$\Rightarrow$x=2009

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