3:
1: \(T=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30\)
=2
2:
ĐKXĐ: 2x-1>=0
=>x>=1/2
\(\sqrt{2x-1}=3x-2\)
=>\(\left\{{}\begin{matrix}3x-2>=0\\\left(3x-2\right)^2=2x-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\9x^2-12x+4-2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\9x^2-14x+5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-1\right)\left(9x-5\right)=0\end{matrix}\right.\Leftrightarrow x=1\)
3:
tan x=4
=>\(\dfrac{sinx}{cosx}=4\)
=>\(sinx=4\cdot cosx\)
\(D=\dfrac{sin^2x-3\cdot sinx\cdot cosx+4\cdot cos^2x}{sin^2x+2\cdot sinx\cdot cosx+3\cdot cos^2x}\)
\(=\dfrac{16\cdot cos^2x-3\cdot4\cdot cosx\cdot cosx+4cos^2x}{16cos^2x+2\cdot4\cdot cosx\cdot cosx+3cos^2x}\)
\(=\dfrac{8cos^2x}{27cos^2x}=\dfrac{8}{27}\)
