\(a>2\\ \Rightarrow\dfrac{1}{a}< \dfrac{1}{2}\\ b>2\\ \Rightarrow\dfrac{1}{b}< \dfrac{1}{2}\\ \dfrac{1}{a}+\dfrac{1}{b}< \dfrac{1}{2}+\dfrac{1}{2}\\ \Rightarrow\dfrac{a+b}{ab}< 1\\ \Rightarrow ab>a+b\)

tuyet qua em oi :3

\(a,đk:\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(TXĐ:D=R/\left\{k\pi;\dfrac{\pi}{2}+k\pi\right\}\)

b, \(y=tanx-cotx=\dfrac{sinx}{cosx}-\dfrac{cosx}{sinx}\)

đk \(\left\{{}\begin{matrix}cosx\ne0\\sinx\ne0\end{matrix}\right.\)

Giống câu trên 

a, Áp dụng định lý Pytago vào tam giác MNP

\(MN^2+MP^2=NP^2\\ \Rightarrow NP=\sqrt{6^2+8^2}=10\left(cm\right)\)

Ta có \(MN\times MP=MI\times NP\\ \Rightarrow MI=\dfrac{6\times8}{10}=4,8\left(cm\right)\)

b, Áp dụng định lý Pytago vào tam giác MNP

\(MN^2+MP^2=NP^2\\ \Rightarrow NP=\sqrt{9^2+16^2}=\sqrt{337}\left(cm\right)\)

Ta cs

\(MN\times MP=MI\times NP\\ \Rightarrow MI=\dfrac{9\times16}{\sqrt{337}}\approx7,8\left(cm\right)\)

c, \(MN^2+MP^2=NP^2\\ \Rightarrow NP=\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2}=\sqrt{5}\left(cm\right)\)

Ta có \(MN\times MP=MI\times NP\\ \Rightarrow MI=\dfrac{\sqrt{2}\times\sqrt{3}}{\sqrt{5}}=\dfrac{\sqrt{30}}{5}\left(cm\right)\)