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S=1/1-1/4+1/4-1/7+.........+1/N-1/N+1
=1/1-(1/4-1/4)+...............+(1/N-1/N)-1/N+1
=1-1/N+1
->S<1
NHA!
\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{\left(3x+1\right).\left(3x+4\right)}\)=\(\dfrac{1344}{2017}\)
\(A=\dfrac{2}{3}(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{3x+1}-\dfrac{1}{3x+4}\))=\(\dfrac{1344}{2017}\)
\(A=\dfrac{2}{3}(1-\dfrac{1}{3x+4})\)=\(\dfrac{1344}{2017}\)
\(A=1-\dfrac{1}{3x+4}=\dfrac{1344}{2017}:\dfrac{2}{3}\)
\(A=1-\dfrac{1}{3x+4}=\dfrac{2016}{2017}\)
\(A=\dfrac{1}{3x+4}=1-\dfrac{2016}{2017}\)
\(A=\dfrac{1}{3x+4}=\dfrac{1}{2017}\)
\(\Rightarrow\)\(3x+4=2017\)
\(3x=2017-4\)
\(3x=2013\)
\(x=671\)
\(\Leftrightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)
\(\rightarrowđpcm\)
Mik cần từ lâu òi , pn trả lời muộn quá !! Nhưng cảm ơn pn na !!!
a)gọi d là ƯCLN (3n-1;6n-3)
\(\Rightarrow\hept{\begin{cases}3n-1⋮d\\6n-3⋮d\end{cases}}\Rightarrow\hept{\begin{cases}6n-2⋮d\\6n-3⋮d\end{cases}}\)
=> (6n-3)-(6n-2)\(⋮\)d
\(\Rightarrow1⋮d\)
=>d=1
\(\Rightarrow\frac{3n-1}{6n-3}\)là pstg(ĐCCM)
b) Gọi d là ƯCLN(2n+11;3n+16)
\(\Rightarrow\hept{\begin{cases}2n+11⋮d\\3n+16⋮d\end{cases}\Rightarrow\hept{\begin{cases}6n+33⋮d\\6n+32⋮d\end{cases}}}\)
\(\Rightarrow\left(6n+33\right)-\left(6n+32\right)⋮d\)
\(\Rightarrow1⋮d\)
=>d=1
Vậy\(\frac{2n+11}{3n+16}\) Là pstg(ĐCCM)
Tớ giải xong rồi ai nhớ nha k cho tôi đi.
\(S=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{n\left(n+3\right)}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{n}-\frac{1}{n+3}\)
\(=1-\frac{1}{n+3}\)
Ta có :
\(\frac{1}{n+3}>0\)
\(\Leftrightarrow-\frac{1}{n+3}< 0\)
\(\Leftrightarrow1-\frac{1}{n+3}< 1\)
\(\Leftrightarrow S< 1\left(đpcm\right)\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n.\left(n+3\right)}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
\(S=1-\frac{1}{n+3}\)
\(S=\frac{n+2}{n+3}\)
Vi \(n\inℕ^∗\)nên \(n+2< n+3\)
DO đó\(\frac{n+2}{n+3}< 1\)
Vậy S <1
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...................+\dfrac{3}{n\left(n+1\right)}\)
\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.............+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow S=1-\dfrac{1}{n+1}< 1\)
\(\Rightarrow S< 1\rightarrowđpcm\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n.\left(n+1\right)}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}\)\(< 1\)
\(\Leftrightarrow S< 1\)
tik cho mik nhé
mình mới học lớp 5
có phải:
E= 1.4+4.7+7.10+...+(3n-2).(3n+1) (với n € N*)
F=2.5+5.8+8.11+...+(3n+2).(3n+5) (với n € N)
G=1.4+7.10+13.16+...+97.100
nếu đúng k cho mình nha