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Câu a đề sai nha bạn
Câu b:
Gọi d=UCLN(21n+4;14n+3)
\(\Leftrightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\Leftrightarrow-1⋮d\)
=>d=1
=>UCLN(42n+8;42n+9)=1
Vậy: 21n+4/14n+3 là phân số tối giản
Câu 1 :\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{100}=\frac{1}{100}\)
Ta có: \(\frac{1}{n}-\frac{1}{n+a}=\frac{1.\left(n+a\right)-1.n}{n\left(n+a\right)}=\frac{n+a-n}{n\left(n+a\right)}=\frac{n-n+a}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
Mà \(\frac{a}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}=>\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}ĐPCM\)
\(\dfrac{1}{k^2}<\dfrac{1}{k(k-1)}=\dfrac{1}{k-1}-\dfrac{1}{k}\)
Ap dung:
\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\ldots+\dfrac{1}{n^2}<1+\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\ldots+\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right)=2-\dfrac{1}{n}<2\)
a) \(\frac{1}{n}\) - \(\frac{1}{n+1}\) = \(\frac{n+1}{n\left(n+1\right)}\) - \(\frac{n}{n\left(n+1\right)}\) = \(\frac{1}{n\left(n+1\right)}\) = \(\frac{1}{n}\) . \(\frac{1}{n+1}\) =>đpcm
b) A= \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{8}\) - \(\frac{1}{9}\) +\(\frac{1}{9}\)
= \(\frac{1}{2}\) + \(\frac{1}{9}\)= \(\frac{11}{18}\)
Ta có: \(\frac{1}{1^2}=\frac{1}{1\cdot1};\frac{1}{2^2}<\frac{1}{1\cdot2};...;\frac{1}{50^2}<\frac{1}{49\cdot50}\)
=>\(\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{50^2}<1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}=1+1-\frac{1}{50}=2-\frac{1}{50}=1,98\)
hay A<1,98 mà 1,98<2 nên A<2
Vậy A<2
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{19.20}<\)\(\frac{1}{2}\)
\(2A<\)\(\frac{1}{2}\)
\(\Rightarrow A<\)\(\frac{1}{4}\)
Vậy \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}<\)\(\frac{1}{4}\)