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=> \(\dfrac{1}{2}.x-\dfrac{1}{3}.x=\dfrac{1}{6}-\dfrac{2}{3}\)
\(x.\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{1}{6}+\dfrac{-4}{6}\)
\(x.\left(\dfrac{3}{6}+\dfrac{-2}{6}\right)=\dfrac{-1}{2}\)
\(x.\dfrac{1}{6}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{1}{6}\)
\(x=\dfrac{-1}{2}.6\)
\(x=-3\)
Vậy x= -3
\(\dfrac{1}{2}x+\dfrac{2}{3}=\dfrac{1}{3}x+\dfrac{1}{6}\)
\(\dfrac{1}{2}x-\dfrac{1}{3}x=\dfrac{1}{6}-\dfrac{2}{3}\)
\(x\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{1}{6}-\dfrac{4}{6}\)
\(x\left(\dfrac{3}{6}-\dfrac{2}{6}\right)=-\dfrac{1}{2}\)
\(x\cdot\dfrac{1}{6}=-\dfrac{1}{2}\)
\(x=-\dfrac{1}{2}:\dfrac{1}{6}\)
\(x=-\dfrac{1}{2}\cdot6\)
\(x=-3\)
Vậy \(x=-3\).
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{\frac{x\left(x+1\right)}{2}}=1\frac{1993}{1991}\)
\(\Leftrightarrow\left(1\cdot\frac{1}{2}\right)+\left(\frac{1}{3}\cdot\frac{1}{2}\right)+\left(\frac{1}{6}\cdot\frac{1}{2}\right)+....+\left(\frac{1}{\frac{x\left(x+1\right)}{2}}\cdot\frac{1}{2}\right)=1\frac{1993}{1991}\div2\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{x\left(x+1\right)}=\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1992}{1991}\)
\(\Leftrightarrow\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{1992}{1991}\)
\(\Leftrightarrow\frac{1}{x+1}=-\frac{1}{1991}\)
\(\Leftrightarrow x=-1992\)
a)\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
b ko hiểu đề
1 + 1/3 + 1/6 + 1/10 + .......... + 1/x.(x+1):2 =1 + 1991/1993
1/2.(1 + 1/3 + 1/6 + 1/10+........+ 1/x.(x+1):2=3984/3986
1/2 + 1/6 +1/12 + .......... +1/x.(x+1)=3984/3986
1/1.2 + 1/2.3 + 1/3.4 +..........+.1/x.(x+1)=3984/3986
2-1/1.2 + 3-2/2.3 + 4-3/3.4 +..........+ x + 1 - x/x.(x+1)
1-1/2+1/2-1/3+1/3-1/4+..........+1/x -1/x+1 =3984/3986
1-1/x+1=3984/3986
1/x+1=1-3984/3986
1/x+1=2/3986=1/1993
x+1=1993
x =1993-1
x =1992
cảm ơn
cảm ơn nhé thế là bai về nhà mình đã giải quyết xong
bạn ai cũng biết tên tôi làm sai rùi'
thank
thank you nhé bạn
Nguyễn Tiến Dũng nói sai thì làm thử đi cho mọi người xem
sai rùi
tìm x ,y,z là nguyên dương sao cho 1 phần y + 1 phần z + 1phan x =2
ai cung biet ten toi làm đúng mà