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

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)

nếu là lớp 8 thì rất hoan nghênh

a) \(\Delta'=\left(\sqrt{3}\right)^2-3\cdot\left(-3\right)=12>0\)

phương trình có 2 nghiệm phân biệt:

\(\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+\sqrt{12}}{3}=\sqrt{3}\\x=\dfrac{\sqrt{3}-\sqrt{12}}{3}=-\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)

kết luận: \(x=\sqrt{3}\), \(x=-\dfrac{\sqrt{3}}{3}\)

b) \(\left\{{}\begin{matrix}x\left(x-1\right)+y=\left(x+1\right)\left(x-3\right)\\2x-3y=-1\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=\left(x+1\right)\left(x-3\right)-x\left(x-1\right)\\5x+9=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

kết luận: \(\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

a) lập delta giải bình thường

b) rút y từ 1 trong 2 pt thế vào pt còn lại

@Leo :) giỏi wa ha . ghen tị ghê

a/ Bạn tự giải

b/ ĐKXĐ:...

Cộng vế với vế: \(\frac{x-y}{y+12}=3\Rightarrow x-y=3y+36\Rightarrow x=4y+36\)

Thay vào pt đầu: \(\frac{4y+36}{y}-\frac{y}{y+12}=1\)
Đặt \(\frac{y+12}{y}=a\Rightarrow4a-\frac{1}{a}=1\Rightarrow4a^2-a-1=0\)

\(\Rightarrow a=\frac{1\pm\sqrt{17}}{8}\) \(\Rightarrow\frac{y+12}{y}=\frac{1\pm\sqrt{17}}{8}\)

\(\Rightarrow\left[{}\begin{matrix}y+12=y\left(\frac{1+\sqrt{17}}{8}\right)\\y+12=y\left(\frac{1-\sqrt{17}}{8}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(\frac{-7+\sqrt{17}}{8}\right)y=12\\\left(\frac{-7-\sqrt{17}}{8}\right)y=12\end{matrix}\right.\) \(\Rightarrow y=...\)

Chắc bạn ghi sai đề, nghiệm quá xấu

3/ \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+y^2=5\\3x^2-9y=3\end{matrix}\right.\) \(\Rightarrow y^2+9y=2\Rightarrow y^2+9y-2=0\Rightarrow y=...\)

4/ ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{3x-1}-3\sqrt{2y+1}=3\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)

\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)

\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)