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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)(nhân mỗi vế với 1/2)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)
\(\Rightarrow x+1=2011\Rightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}\right)=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)\(=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)\(=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Rightarrow x+1=2011\)
\(\Rightarrow x=2010\)
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x/4=18/x+1
=>x(x+1)=4.18=72=8.9=8.(8+1)
hoặc x(x+1)=72=(-9).(-8)=(-9).[(-9)+1]
=>x=8 hoặc x=-9
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\(\frac{-1}{3}\le x\le\frac{-2}{7}\Rightarrow\frac{-7}{21}\le x\le\frac{-6}{21}\Rightarrow x\varepsilon\left\{\frac{-7}{21};\frac{-6}{21}\right\}\)
\(\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow\)\(x\left(x+1\right)=4.18\)
\(\Leftrightarrow\)\(x\left(x+1\right)=72\)
\(\Leftrightarrow\)\(x^2+x-72=0\)
\(\Leftrightarrow\)\(\left(x-8\right)\left(x+9\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-8=0\\x+9=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=8\\x=-9\end{cases}}\)
Vậy....
tim x thuoc Z biet x4 = 18x+1
giup voi
Toán lớp 6
Thống kê hỏi đáp
Báo cáo sai phạm
x4 =18x+1
⇔x(x+1)=4.18
⇔x(x+1)=72
⇔x2+x−72=0
⇔(x−8)(x+9)=0
⇔[
⇔[
Vậy \(x=8\)hoặc \(-9\)