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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
⇒ 1/k² < 1/[(k-1).(k+1)] = [1/(k-1) - 1/(k+1)]/2 (*)
Áp dụng (*), ta có:
1/2² + 1/3² + 1/4² + ... + 1/n²
< 1/2² + 1/(2.4) + 1/(3.5) + ... + 1/[(n-1).(n+1)]
= 1/2² + [1/2 - 1/4 + 1/3 - 1/5 + ... + 1/(n-1) - 1/(n+1)]/2
= 1/2² + [1/2 + 1/3 - 1/n - 1/(n+1)]/2
= 2/3 - [1/n + 1/(n+1)]/2 <2/3<1
Vì:\(\dfrac{1}{2^{^2}}>\dfrac{1}{1.2};\dfrac{1}{3^{^2}}>\dfrac{1}{2.3};\dfrac{1}{4^{^2}}>\dfrac{1}{3.4};...;\dfrac{1}{n^{^2}}>\dfrac{1}{\left(n-1\right)n}\)
\(\Rightarrow A>\dfrac{1 }{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\Rightarrow A>\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow A>\dfrac{1}{1}-\dfrac{1}{n}\)
....
Tự lập luận làm tiếp
Ta có:\(\dfrac{1}{2^3}< \dfrac{1}{1.2.3};\dfrac{1}{3^3}< \dfrac{1}{2.3.4};\dfrac{1}{4^3}< \dfrac{1}{3.4.5};...;\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)Vậy:\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+...+\dfrac{1}{n^3}< \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)Ta có:\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)
=\(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n.\left(n+1\right)}\right)\)=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{n.\left(n+1\right)}\right)\)
=\(\dfrac{1}{4}-\dfrac{1}{2n.\left(n+1\right)}< \dfrac{1}{4}\)
Vì:\(\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}-\dfrac{1}{2n.\left(n+1\right)}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{2^3}+\dfrac{1}{3^3}+...+\dfrac{1}{n^3}< \dfrac{1}{4}\) hay \(A< \dfrac{1}{4}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}\\ A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\left(\dfrac{1}{n}>0\right)\)