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

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

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

Bạn này ngầu quá ạ 

\(5^{x+1}-5^x=2.2^x+8.2^x\\ \Leftrightarrow5^x.5-5^x=2.2^x+8.2^x\\ \Leftrightarrow5^x\left(5-1\right)=2^x\left(2+8\right)\\ \Leftrightarrow5^x.4=2^x.10\\ \Leftrightarrow5^x:2^x=\dfrac{5}{2}\\ \Leftrightarrow\left(\dfrac{5}{2}\right)^x=\dfrac{5}{2}\\ \Rightarrow x=1\)

\(a,,0< x< \dfrac{\pi}{2}\\ \Rightarrow\left\{{}\begin{matrix}sinx>0\\cosx< 0\end{matrix}\right.\\ 1+tan^2x=\dfrac{1}{cos^2x}\\ \Rightarrow cos^2x=\dfrac{1}{4}\\ \Rightarrow cosx=-\dfrac{1}{2}\)

\(sin^2x+cos^2x=1\\ \Rightarrow sin^2x=1-\left(-\dfrac{1}{2}\right)^2\\ =\dfrac{3}{4}\\ \Rightarrow sinx=\dfrac{\sqrt{3}}{2}\)

\(tanx.cotx=1\\ \Rightarrow cotx=1:\sqrt{3}\\ =\dfrac{\sqrt{3}}{3}\)

\(b,\dfrac{3\pi}{2}< x< 2\pi\\ \Rightarrow\left\{{}\begin{matrix}sinx< 0\\cosx>0\end{matrix}\right.\)

\(tanx.cotx=1\\ \Rightarrow tanx=-1\)

\(1+cot^2x=\dfrac{1}{sin^2x}\\ \Rightarrow sin^2x=\dfrac{1}{2}\\ \Rightarrow sinx=-\dfrac{\sqrt{2}}{2}\\ cos^2x+sin^2x=1\\ \Rightarrow cos^2x=\dfrac{1}{2}\\ \Rightarrow cosx=\dfrac{\sqrt{2}}{2}\)