Mình mới lớp 5 nên không biết làm bài này.
Xin lỗi nha! Chúc bạn may mắn......mình chính là Đào Minh Tiến!
a) \(\frac{n}{n+1}\)và \(\frac{n+2}{n+3}\)
\(\frac{n}{n+1}=\frac{n\cdot\left(n+3\right)}{\left(n+1\right)\cdot\left(n+3\right)}\)
\(\frac{n+2}{n+3}=\frac{\left(n+2\right)\cdot\left(n+1\right)}{\left(n+3\right)\cdot\left(n+1\right)}\)
So sánh : \(n\cdot\left(n+3\right)\)và \(\left(n+2\right)\cdot\left(n+3\right)\)
\(n\cdot\left(n+3\right)=n^2+3n\)
\(\left(n+2\right)\cdot\left(n+3\right)=n^2+5n+6\)
\(n^2+3n< n^2+5n+6\)
\(\Leftrightarrow\frac{n}{n+1}< \frac{n+2}{n+3}\)
b) \(\frac{n}{2n+1}\)và \(\frac{3n+1}{6n+3}\)
\(\frac{n}{2n+1}=\frac{n\cdot\left(6n+3\right)}{\left(2n+1\right)\cdot\left(6n+3\right)}\)
\(\frac{3n+1}{6n+3}=\frac{\left(3n+1\right)\cdot\left(2n+1\right)}{\left(6n+3\right)\cdot\left(2n+1\right)}\)
So sánh : \(n\cdot\left(6n+3\right)\)và \(\left(3n+1\right)\cdot\left(2n+1\right)\)
\(n\cdot\left(6n+3\right)=6n^2+3n\)
\(\left(3n+1\right)\cdot\left(2n+1\right)=6n^2+5n+1\)
\(6n^2+3n< 6n^2+5n+1\)
\(\Leftrightarrow\frac{n}{2n+1}< \frac{3n+1}{6n+3}\)
a) Trả lời:
Vì \(\frac{n}{n+1}< 1\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+1+2}\)
\(\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+3}\)
Câu này dựa vào định nghĩa:
Nếu \(\frac{a}{b}< 1\) thì \(\frac{a}{b}< \frac{a+m}{b+m}\).
