a, Áp dụng HTL: \(\left\{{}\begin{matrix}y=\sqrt{3\left(3+12\right)}=3\sqrt{5}\\z=\sqrt{12\left(3+12\right)}=6\sqrt{5}\\x=\sqrt{3\cdot12}=6\end{matrix}\right.\)
b, \(x=\sqrt{8^2+6^2}=10\left(pytago\right)\)
Áp dụng HTL: \(\left\{{}\begin{matrix}z=\dfrac{6^2}{10}=3,6\\t=\dfrac{8^2}{10}=6,4\\y=\dfrac{8\cdot6}{10}=4,8\end{matrix}\right.\)
a) \(x^2=3\cdot12=36\Rightarrow x=6\)
\(y^2=3\cdot\left(3+12\right)=45\Rightarrow y=3\sqrt{5}\)
\(z^2=12\cdot\left(3+12\right)=180\Rightarrow z=6\sqrt{5}\)
b) \(x^2=6^2+8^2=100\Rightarrow x=10\)
\(6^2=z\cdot x=z\cdot10\Rightarrow z=3,6\)
\(8^2=t\cdot x=t\cdot10\Rightarrow t=6,4\)
\(y^2=z\cdot t=3,6\cdot6,4=23,04\Rightarrow y=4,8\)
