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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2042}\)
<=> \(\frac{1}{x+1}=\frac{1}{2021}\)
<=> x + 1 = 2021
<=> x = 2020
Có phải là bình 6a3 học trường THCS Nguyễn Trãi đúng không
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\(\Leftrightarrow1+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{x\left(x+1\right)}=1+\dfrac{2019}{2021}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2019}{2021}\)
\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2019}{2021}\)
\(\Leftrightarrow\dfrac{2}{x+1}=1-\dfrac{2019}{2021}\)
\(\Leftrightarrow\dfrac{2}{x+1}=\dfrac{2}{2021}\)
\(\Leftrightarrow x+1=2021\)
\(\Leftrightarrow x=2020\)
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Đề bạn thiếu 1 số \(x\) nữa đúng không?
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2021}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2021}\)
\(\Rightarrow x+1=2021\)
\(\Rightarrow x=2020\)
Vậy \(x=2020\).
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2021}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4042}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4042}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2019}{4042}=\frac{1}{2021}\)
\(\Leftrightarrow x+1=2021\)
\(\Leftrightarrow x=2020\left(tm:x\in N\right)\)
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Bài 2:
Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)
\(\Leftrightarrow16x+40=90+30\)
\(\Leftrightarrow16x=80\)
hay x=5
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\(x_1+x_2=x_3+x_4=...=x_{2019}+x_{2020}=2\Rightarrow x_1+x_2+x_3+x_4+...+x_{2019}+x_{2020}=2.1010=2020\)
\(\Rightarrow x_1+x_2+x_3+x_4+...+x_{2019}+x_{2020}+x_{2021}=2020+x_{2021}\)
\(\Rightarrow0=2020+x_{2021}\)
\(\Rightarrow x_{2021}=-2020\)
Vậy \(x_{2021}=-2020\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}\)
\(=2-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{x}-\frac{2}{x+1}\)
\(=2-\frac{2}{x+1}\)
Do đó phương trình ban đầu tương đương với:
\(2-\frac{2}{x+1}=1\frac{2019}{2021}\)
\(\Leftrightarrow x=2019\).