thang ham

de vay

\(\Leftrightarrow2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(\Leftrightarrow\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2010}{4024}=\dfrac{1005}{2012}\)

=>1/x+1=-251/1006

=>x+1=-1006/251

=>x=-1257/251

\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{2}{x\left(x+1\right)}=\frac{2010}{2012}\)

\(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{x\left(x+1\right)}=\frac{2010}{2012}\)

\(\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{x\left(x+1\right)}=\frac{2010}{2012}\)

\(2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2010}{2012}\)

\(2\left(\frac{1}{4}-\frac{1}{x+1}\right)=\frac{2010}{2012}\)

\(\frac{1}{4}-\frac{1}{x+1}=\frac{2010}{2012}\div2\)

\(\frac{1}{4}-\frac{1}{x+1}=\frac{1005}{2012}\)

\(\frac{1}{x+1}=\frac{1}{4}-\frac{1005}{2012}\)

\(\frac{1}{x+1}=\frac{-502}{2012}=-\frac{251}{1006}\)

\(\Rightarrow x+1=1\div-\frac{251}{1006}=-\frac{1006}{251}\)

\(x=\frac{-1006}{251}-1=-\frac{1257}{251}\)

bản rút gọn 2/x(x+1) thanh 1/x(x+1) luon di

\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)

\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)

\(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)

\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)

\(\Rightarrow\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)

\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)

\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)

\(\left(x+1\right)=1:\dfrac{-251}{1006}\)

\(x+1=\dfrac{-1006}{251}\)

\(x=\dfrac{-1006}{251}-1\)

\(x=\dfrac{-1257}{251}\)

Nếu bạn tìm \(x\in Z\) hay \(x\in N\) thì \(x=\varnothing\) (không có x thoả mãn)

gì thế

lớp 6 đây

á...á

cưus

k mk nha

https://hoc24.vn/question/246430.html

\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)

\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{2010}{2012}:2\)

\(\dfrac{1}{4}-\dfrac{1}{\left(x+1\right)}=\dfrac{1005}{2012}\)

\(\dfrac{1}{\left(x+1\right)}=\dfrac{1}{4}-\dfrac{1005}{2012}\)

\(\dfrac{1}{\left(x+1\right)}=\dfrac{-251}{1006}\)

\(\Rightarrow1:\left(x+1\right)=\dfrac{-251}{1006}\)

\(\left(x+1\right)=1:\dfrac{-251}{1006}\)

\(x+1=\dfrac{-1006}{251}\)

\(x=\dfrac{-1006}{251}-1\)

\(x=\dfrac{-1257}{251}\)

Vì \(x\in N\) nên \(x=\varnothing\) (không có giá trị nào của x thoả mãn)

Ta có :

\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+.........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2010}{2012}\)

\(\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+..........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2010}{2012}\)

\(\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{7.8}+.......+\dfrac{2}{x\left(x+1\right)}=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{7.8}+...........+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.........+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2010}{2012}\)

\(2\left(\dfrac{1}{4}-\dfrac{1}{x+1}\right)=\dfrac{2010}{2012}\)

\(\dfrac{1}{4}-\dfrac{1}{x+1}=\dfrac{2010}{2012}:2\)

\(\dfrac{1}{4}-\dfrac{1}{x+1}=\dfrac{1005}{2012}\)

\(\dfrac{1}{x+1}=\dfrac{1}{4}-\dfrac{1005}{2012}\)

\(\dfrac{1}{x+1}=\dfrac{-215}{1006}\)

\(\Rightarrow1.1006=\left(x+1\right).\left(-215\right)\)

\(1006=\left(x+1\right).\left(-215\right)\)

\(x+1=1006:\left(-215\right)\)

\(x+1=\dfrac{-1006}{215}\)

\(x=\dfrac{-1006}{215}-1\)

\(x=\dfrac{-1221}{215}\)(ko thỏa mãn \(x\in N\))

Vậy ko tìm dc giá trị của x thỏa mãn theo yêu cầu

~ Học tốt ~