a) Ta có: \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-4\left|y\right|=18\\6x+9\left|y\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13\left|y\right|=15\\3x-2\left|y\right|=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=\dfrac{-15}{13}\\3x-2\left|y\right|=9\end{matrix}\right.\Leftrightarrow\)Phương trình vô nghiệmVậy: \(S=\varnothing\)

$\begin{cases}3x-2|y|=9\\2x+3|y|=1\\\end{cases}$

`<=>` $\begin{cases}6x-4|y|=18\\6x+9|y|=3\\\end{cases}$

`<=>` $\begin{cases}13|y|=-15(loại)\\|3x|-2|y|=9\\\end{cases}$

Vậy HPT vô nghiệm

$\begin{cases}|x-2|+2|y-1|=9\\x+|y-1|=-1\\\end{cases}$

`<=>` $\begin{cases}|x-2|+2|y-1|=9\\2x+2|y-1|=-2\\\end{cases}$

`<=>` $\begin{cases}|x-2|-2x=11\\x+|y-1|=-1\\\end{cases}$

`<=>` $\begin{cases}|x-2|=2x+11\\x+|y-1|=-1\\\end{cases}$

Đến đây dễ rồi bạn tự giải :D

nhân pt (2) vs 3 sau đó cộng pt (1) vs (2) ta đc

\(\left\{{}\begin{matrix}x^3+3xy^2=-46\\x^3+3xy^2+3x^2-24xy+3y^2=24y-51x-46\end{matrix}\right.\)

bây h ta chú ý tới pt dưới

\(x^3+3xy^2+3x^2-24xy+3y^2-24y+51x+46=0\)

\(\left(x+1\right)\left(x^2+2x+3y^2-24y+49\right)=0\)

\(\left(x+1\right)\left[\left(x+1\right)^2+3\left(y-4\right)^2\right]=0\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\x^3+3xy^2=-49\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\end{matrix}\right.\rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\end{matrix}\right.\)

vậy hệ có 2 nghiệm

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

\(2x^2-\left(3y-3\right)x+y^2-2y+1=0\)

\(\Delta=\left(3y-3\right)^2-8\left(y^2-1y+1\right)=\left(y-1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3y-3+y-1}{4}\\x=\dfrac{3y-3-y+1}{4}\end{matrix}\right.\)

\(\Rightarrow...\)

\(\left\{{}\begin{matrix}9x^2-3xy+2y^2=23\\7x^2+6xy-8y^2=-37\end{matrix}\right.\)\(\left(hpt\right)\)

\(đặt:x=t.y\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}9\left(t.y\right)^2-3t.y^2+2y^2=23\left(1\right)\\7\left(ty\right)^2+6t.y^2-8y^2=-37\left(2\right)\end{matrix}\right.\)

\(\Rightarrow-37\left[9\left(t.y\right)^2-3ty^2+2y^2\right]=23\left[7\left(ty\right)^2+6ty^2-8y^2\right]\)

\(\Leftrightarrow494\left(ty\right)^2+27ty^2-110y^2=0\left(3\right)\)

\(x=y=0\) \(không\) \(là\) \(nghiệm\) \(hpt\)

\(y\ne0\Rightarrow\left(3\right)\Leftrightarrow494t^2+27t-110=0\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{110}{247}\Rightarrow x=\dfrac{110}{247}.y\left(4\right)\\t=-\dfrac{1}{2}\Rightarrow x=-\dfrac{1}{2}.y\left(5\right)\end{matrix}\right.\)

\(thay\left(4\right)và\left(5\right)vào-hpt\Rightarrow x,y=.....\)(đến đây dễ rồi bạn tự tìm x,y)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+3xy^2=-49\\3x^2-24xy+3y^2=24y-51x\end{matrix}\right.\)

Cộng vế:

\(x^3+3x^2+3y^2\left(x+1\right)-24y\left(x+1\right)+51x+49=0\)

\(\Leftrightarrow\left(x+1\right)^3+3y^2\left(x+1\right)-24y\left(x+1\right)+48\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^3+3\left(x+1\right)\left(y-4\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2+3\left(y-2\right)^2\right]=0\)