1:
a: \(A=2\sqrt{5}-3\sqrt{5}+3\cdot3\sqrt{2}+6\sqrt{2}\)
\(=-\sqrt{5}+15\sqrt{2}\)
b: \(B=0.1\cdot10\sqrt{2}+2\cdot\dfrac{1}{5}\sqrt{2}+0.4\cdot2\sqrt{5}\)
\(=\sqrt{2}+\dfrac{2}{5}\sqrt{2}+0.8\sqrt{5}=\dfrac{7}{5}\sqrt{2}+\dfrac{4}{5}\sqrt{5}\)
2:
1: x^2-4x-5=0
=>x^2-5x+x-5=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
2: \(\left\{{}\begin{matrix}7x+4y=74\\3x+2y=32\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x+4y=74\\6x+4y=64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\3x+2y=32\end{matrix}\right.\)
=>x=10 và y=1
3:
x-3y=6 và -2x+6y=-12
=>x-3y=6 và -x+3y=-6
=>\(\left(x,y\right)\in R\)
Vậy: \(\left\{{}\begin{matrix}y\in R\\x=3y+6\end{matrix}\right.\)
4: \(3x^2-2\sqrt{3}x-3=0\)
=>\(3x^2-3\sqrt{3}x+\sqrt{3}x-3=0\)
\(\Leftrightarrow3x\left(x-\sqrt{3}\right)+\sqrt{3}\left(x-\sqrt{3}\right)=0\)
=>\(\left(x-\sqrt{3}\right)\left(3x+\sqrt{3}\right)=0\)
=>\(\left[{}\begin{matrix}x=\sqrt{3}\\x=-\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
