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\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\dfrac{1}{3}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\dfrac{-1}{\left(n+1\right)}=\dfrac{-1}{30}\)
\(-n-1=-30\)
-n = -29
n = 29
\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{3}-\dfrac{1}{n+1}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{1}{n+1}=\dfrac{1}{30}\)
\(\Rightarrow n+1=30\)
\(\Rightarrow n=29\)
Vậy n = 29.
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(A=1-\frac{1}{6}=\frac{5}{6}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(B=1-\frac{1}{n+1}=\frac{n}{n+1}\)
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{30}\)
\(\Rightarrow x=30-1\)
\(\Rightarrow x=29\)
vậy: \(x=29\)
(1/3-1/4+1/4-1/5+1/5-.......+1/x.(x+1)=3/10
1/3-1/x+1=3/10
tự làm...
\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + .....+\(\dfrac{1}{n.(n+1)}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\) +......+ \(\dfrac{1}{n}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{3}-\dfrac{1}{n+1}\) = \(\dfrac{3}{10}\)
\(\dfrac{1}{n+1}\) = \(\dfrac{1}{3}-\dfrac{3}{10}\)
\(\dfrac{1}{n+1}\) = \(\dfrac{1}{30}\)
n + 1 = 30
n = 30 - 1
n = 29
Kết luận n = 29 là giá trị thỏa mãn yêu cầu đề bài.