a)

∆'=3+9=12

x1=(√3-2√3)/3=-√3/3

x2=(√3+2√3)/3=√3

b.

<=>

x+y=-3(1)

2x-3y=-1(2)

(1).2-(2)<=>5y=-5;y=-1

=>(x,y)=(-2;-1)

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

Xem lại giúp tớ dấu căn ở câu c và d nhé.  

a) Ta có: \(\left\{{}\begin{matrix}-x+2y=3\\3x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=9\\3x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=8\\-x+2y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{8}{7}\\-x=3-2y=3-2\cdot\dfrac{8}{7}=\dfrac{5}{7}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}2x+2\sqrt{3}\cdot y=1\\\sqrt{3}x+2y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3}x+6y=\sqrt{3}\\2\sqrt{3}x+4y=-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y=\sqrt{3}+10\\\sqrt{3}x+2y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}+2\cdot\dfrac{\sqrt{3}+10}{2}=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}=-5-\sqrt{3}-10=-15-\sqrt{3}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)

a, \(\left\{{}\begin{matrix}\\6x+2y=-2\end{matrix}\right.-6x+12y=18}\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\) 

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)

b.

ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)

Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:

\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)

\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)

\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)

Thay xuống pt dưới:

\(6y+y=14\Rightarrow y=2\)

\(\Rightarrow x=4\)

\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)