Question

[Đỗ Thị Minh Anh]

so sánh 2 phân số

n/n+2 và n-1/n+4

Answer 1

[Đinh Tuấn Việt]

\(1-\frac{n}{n+2}=\frac{n+2}{n+2}-\frac{n}{n+2}=\frac{2}{n+2}\)

\(1-\frac{n-1}{n+4}=\frac{n+4}{n+4}-\frac{n-1}{n+4}=\frac{n+5}{n+4}\)

Mà \(\frac{2}{n+2}<1\) và \(\frac{n+5}{n+4}>1\)nên \(\frac{2}{n+2}<\frac{n+5}{n+4}\).

Do đó \(\frac{n}{n+2}<\frac{n-1}{n+4}\)

Answer 2

[Lê Song Thanh Nhã]

\(\frac{n}{n+2}=\frac{n+2-2}{n+2}=\frac{n+2}{n+2}-\frac{2}{n+2}=1-\frac{2}{n+2}\)

\(\frac{n-1}{n+4}=\frac{n+4-5}{n+4}=\frac{n+4}{n+4}-\frac{5}{n+4}=1-\frac{5}{n+4}\)

Ta có: \(\frac{2}{n+2}=\frac{\left(n+4\right)2}{\left(n+4\right)\left(n+2\right)}=\frac{2n+8}{n^2+2n+4n+8}\)

\(\frac{5}{n+4}=\frac{\left(n+2\right)5}{\left(n+2\right)\left(n+4\right)}=\frac{5n+10}{n^2+4n+2n+8}\)

Vì \(\frac{2n+8}{n^2+2n+4n+8}<\frac{5n+10}{n^2+4n+2n+8}\)

Nên \(1-\frac{5}{n+4}<1-\frac{2}{n+2}\)

Vậy \(\frac{n}{n+2}>\frac{n-1}{n+4}\)

ĐÚNG NHAAA. Mìh mất côq giải đó