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Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n-1\right)}\)
\(\Rightarrow P< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow P< 1-\dfrac{1}{n}< 1\)
\(\Rightarrow P< 1\)
Bài 1:
\(\left|5x-4\right|=\left|x+2\right|\)
\(\Rightarrow\left\{{}\begin{matrix}5x-4=-\left(x+2\right)\\5x-4=x+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x-4=-x-2\\5x-x=2+4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5x+x=-2+4\\4x=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}6x=2\\x=\dfrac{6}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{3};\dfrac{3}{2}\right\}\)
Bài 2:
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};.....;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right).n}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right).n}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1}-\dfrac{1}{n}< 1\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1\)
Chúc bạn học tốt nha!!!
\(S_n=1-\dfrac{1}{n^2}\) xét tổng \(U_n=\dfrac{1}{n^2}\) với n >=2
cơ bản có \(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}=\dfrac{1}{n-1}-\dfrac{1}{n}\)
<=>\(U< 1-\dfrac{1}{n-1}\)
cơ bản có \(\dfrac{1}{n^2}>\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
<=>\(U>1-\dfrac{1}{n+1}\)
<=>\(1-\dfrac{1}{n-1}< U< 1-\dfrac{1}{n+1}\)
với n >2 => 1/(n-1) ; 1/(n+1) là hai phân số <1
=> U không phải là số nguyên
=> S không là số nguyên => dpcm
Ta thấy:
\(1\cdot2^2=2^2;2\cdot3^2>3^2;3\cdot4^2>4^2;...;49\cdot50^2>50^2\)
\(\Rightarrow\dfrac{1}{1.2^2}=\dfrac{1}{2^2};\dfrac{1}{2\cdot3^2}< \dfrac{1}{3^2};\dfrac{1}{3\cdot4^2}< \dfrac{1}{4^2};...;\dfrac{1}{49\cdot50^2}< \dfrac{1}{50^2}\)
\(\Rightarrow\dfrac{1}{1\cdot2^2}+\dfrac{1}{2\cdot3^2}+\dfrac{1}{3\cdot4^2}+...+\dfrac{1}{49\cdot50^2}< \dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
hay A<B
Vậy A<B
a/ Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+.........+\dfrac{1}{3^{50}}\)
\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.......+\dfrac{1}{3^{49}}\)
\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+....+\dfrac{1}{3^{49}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{50}}\right)\)
\(\Leftrightarrow2A=1-\dfrac{1}{3^{50}}\)
còn sao nx thì mk chịu =.=
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}\\ A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\left(\dfrac{1}{n}>0\right)\)