Tìm x biết 1 + 1/3 + 1/6 + 1/10+ ................+ 1/ x ( x +1 ) :2 = 1 + 1991/1993
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Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
Bài 2:
h; \(\dfrac{2}{3}\)\(x\) + 50% + \(x\) = \(\dfrac{1}{10}\)
\(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\) + \(x\) = \(\dfrac{1}{10}\)
(\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)
\(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)
\(x\) = \(\dfrac{-2}{5}\): \(\dfrac{5}{3}\)
\(x\) = - \(\dfrac{6}{25}\)
Lớp 5 chưa học số âm em nhé.
=> \(\dfrac{1}{2}.x-\dfrac{1}{3}.x=\dfrac{1}{6}-\dfrac{2}{3}\)
\(x.\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{1}{6}+\dfrac{-4}{6}\)
\(x.\left(\dfrac{3}{6}+\dfrac{-2}{6}\right)=\dfrac{-1}{2}\)
\(x.\dfrac{1}{6}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}:\dfrac{1}{6}\)
\(x=\dfrac{-1}{2}.6\)
\(x=-3\)
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 + 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
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