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\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)
\(\Rightarrow\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)
\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)
\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)
\(\left(x+1\right)=1:\dfrac{-251}{1006}\)
\(x+1=\dfrac{-1006}{251}\)
\(x=\dfrac{-1006}{251}-1\)
\(x=\dfrac{-1257}{251}\)
Nếu bạn tìm \(x\in Z\) hay \(x\in N\) thì \(x=\varnothing\) (không có x thoả mãn)
\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)
\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)
\(\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)
\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)
\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)
\(\left(x+1\right)=1:\dfrac{-251}{1006}\)
\(x+1=\dfrac{-1006}{251}\)
\(x=\dfrac{-1006}{251}-1\)
\(x=\dfrac{-1257}{251}\)
Vì \(x\in N\) nên \(x=\varnothing\) (không có giá trị nào của x thoả mãn)
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
a)\(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=\dfrac{-5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
b) \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
a) Ta có: \(\left|x+\dfrac{2}{3}\right|=\dfrac{5}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=-\dfrac{5}{6}\\x+\dfrac{2}{3}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)
b) Ta có: \(\left(x-\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{2}{3}\\x-\dfrac{1}{3}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
\(=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}-\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{115}{-161}=-\dfrac{115}{161}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}\)
\(A-\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)
\(\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)
\(A=\dfrac{1}{2^{2022}}-2\)
12A=12+(12)2+(12)3+(12)4+...+(12)202312A=12+(12)2+(12)3+(12)4+...+(12)2023
A−12A=(12)2023−1A−12A=(12)2023−1
12A=(12)2023−112A=(12)2023−1
A=122022−2
a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)
b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)
c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)
\(\Leftrightarrow2\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1005}{1006}\)
\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1005}{2012}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{-251}{1006}\)
=>x+1=-1006/251
hay x=-1257/251