Câu 30:
\(\dfrac{1}{101}< \dfrac{1}{100}\)
\(\dfrac{1}{102}< \dfrac{1}{100}\)
...
\(\dfrac{1}{200}< \dfrac{1}{100}\)
Do đó: \(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{100}{100}=1\)
\(\dfrac{1}{101}>\dfrac{1}{200}\)
\(\dfrac{1}{102}>\dfrac{1}{200}\)
...
\(\dfrac{1}{200}=\dfrac{1}{200}\)
Do đó: \(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)
Do đó: \(\dfrac{1}{2}< A< 1\)
Câu 27:
a: Gọi d=ƯCLN(n+2;2n+5)
=>\(\left\{{}\begin{matrix}n+2⋮d\\2n+5⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2n+4⋮d\\2n+5⋮d\end{matrix}\right.\)
=>\(2n+4-2n-5⋮d\)
=>\(-1⋮d\)
=>d=1
=>ƯCLN(n+2;2n+5)=1
=>\(\dfrac{n+2}{2n+5}\) là phân số tối giản
b: Gọi d=ƯCLN(2n+3;4n+8)
=>\(\left\{{}\begin{matrix}2n+3⋮d\\4n+8⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4n+8⋮d\\4n+6⋮d\end{matrix}\right.\)
=>\(4n+8-4n-6⋮d\)
=>\(2⋮d\)
mà 2n+3 lẻ
nên d=1
=>ƯCLN(2n+3;4n+8)=1
=>\(\dfrac{2n+3}{4n+8}\) là phân số tối giản
c: Câu này sai với n=3 nha bạn
