1) Mình vừa làm rồi zô trang cá nhân xem
2) Ta có: \(\left\{{}\begin{matrix}a>2\\b>2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2+m\\b=2+n\end{matrix}\right.\)
\(ab=\left(2+m\right)\left(2+n\right)=4+2n+2m+mn\)
\(a+b=2+m+2+n=4+m+n\)
Dễ thấy: \(4+2\left(m+n\right)>4+m+n\)
Nên \(4+2m+2n+mn>4+m+n\)
Hay \(ab>a+b\left(đpcm\right)\)
4)
Vì \(0< a_1< a_2< ...< a_{15}\) nên:
\(\dfrac{a_1+a_2+a_3+...+a_{15}}{a_5+a_{10}+a_{15}}< \dfrac{a_5+a_{10}+a_{15}+a_5+a_{10}+a_{15}+...+a_5+a_{10}+a_{15}}{a_5+a_{10}+a_{15}}=\dfrac{5\left(a_5+a_{10}+a_{15}\right)}{a_5+a_{10}+a_{15}}=5\left(đpcm\right)\)Đề lớp 6 ak?
Đúng 0
Bình luận (0)
