Điều kiện: \(y\ge0\)

pt thứ nhất của hệ \(\Leftrightarrow\left(y-x+3\right)^2=0\) \(\Leftrightarrow y-x+3=0\) \(\Leftrightarrow y=x-3\)

Thay vào pt thứ hai của hệ, ta được  \(2x^2+3x+x-3-\left(3x+1\right)\sqrt{x-3}-2=0\)

\(\Leftrightarrow2x^2+4x-5=\left(3x+1\right)\sqrt{x-3}\)         \(\left(x\ge3\right)\)

\(\Rightarrow\left(2x^2+4x-5\right)^2=\left[\left(3x+1\right)\sqrt{x-3}\right]^2\)

\(\Leftrightarrow4x^4+16x^2+25+16x^3-20x^2-40x=\left(3x+1\right)^2\left(x-3\right)\)

\(\Leftrightarrow4x^4+16x^3-4x^2-40x+25=9x^3-21x^2-17x-3\)

\(\Leftrightarrow4x^4+7x^3+17x^2-23x+28=0\)

Đặt \(f\left(x\right)=4x^4+7x^3+17x^2-23x+28\)

\(f\left(x\right)=4x^4+7x^3+17x^2+4+4+...+4-23x+4\) (có 6 số 4 ở giữa)

\(f\left(x\right)\ge9\sqrt[9]{4x^4.7x^3.17x^2.4^6}-23x+4\) \(=\left(9\sqrt[9]{1949696}-23\right)x+4\)

Hiển nhiên \(9\sqrt[9]{1949696}>23\). Lại có \(x\ge3\) nên \(f\left(x\right)>0\), Như vậy pt \(f\left(x\right)=0\) vô nghiệm. Điều đó có nghĩa là phương trình đã cho vô nghiệm.

Lời giải:

$3x^2-4xy+y^2=0$

$\Leftrightarrow 3x(x-y)-y(x-y)=0$

$\Leftrightarrow (x-y)(3x-y)=0$
$\Rightarrow x-y=0$ hoặc $3x-y=0$

Nếu $x-y=0\Leftrightarrow x=y$. Thay vào pt $(2)$:
$x^2+2x=8$

$\Leftrightarrow x^2+2x-8=0$

$\Leftrightarrow (x-2)(x+4)=0$

$\Rightarrow x=2$ hoặc $x=-4$. 

Vậy hpt có nghiệm $(x,y)=(2,2); (-4,-4)$

Nếu $3x-y=0$

$\Leftrightarrow 3x=y$. Thay vô pt $(2)$:

$x^2+6x=8$

$\Leftrightarrow x^2+6x-8=0$
$\Rightarrow x=-3\pm \sqrt{17}$

$\Rightarrow y=3(-3\pm \sqrt{17})$ (tương ứng) 

Vậy tổng cộng hpt có 4 nghiệm $(x,y)$ thực.

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:

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

\(\Rightarrow x=4\)

\(\left\{{}\begin{matrix}\left(x+1\right)^2-y^2+6y-9=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y^2-6y+9\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y-3\right)^2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+1+y-3\right)\left(x+1-y+3\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+y-2\right)\left(x-y+4\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)

TH1: x+y-2=0

=>x=-y+2

\(2x^2+y^2-6y-1=0\)

=>\(2\left(-y+2\right)^2+y^2-6y-1=0\)

=>\(2\left(y^2-4y+4\right)+y^2-6y-1=0\)

=>\(3y^2-14y+7=0\)

\(\Delta=\left(-14\right)^2-2\cdot3\cdot7=196-42=154>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}y=\dfrac{14-\sqrt{154}}{6}\\y=\dfrac{14+\sqrt{154}}{6}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-y+2=\dfrac{-2+\sqrt{154}}{6}\\x=\dfrac{-2-\sqrt{154}}{6}\end{matrix}\right.\)

TH2: x-y+4=0

=>x=y-4

\(2x^2+y^2-6y-1=0\)

=>\(2\left(y-4\right)^2+y^2-6y-1=0\)

=>\(2\left(y^2-8y+16\right)+y^2-6y-1=0\)

=>\(3y^2-22y+31=0\)

\(\Delta=\left(-22\right)^2-4\cdot3\cdot31=112>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}y_1=\dfrac{22-\sqrt{112}}{2\cdot3}=\dfrac{11-\sqrt{28}}{3}\\y_2=\dfrac{22+\sqrt{112}}{2\cdot3}=\dfrac{11+\sqrt{28}}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=y-4=\dfrac{11-\sqrt{28}}{3}-4=\dfrac{-1-\sqrt{28}}{3}\\x=y-4=\dfrac{11+\sqrt{28}}{3}-4=\dfrac{-1+\sqrt{28}}{3}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(x+1\right)^2-y^2+6y-9=0\\2x^2+y^2-6y+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y^2-6y+9\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2-\left(y-3\right)^2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y-2\right)\left(x-y+4\right)=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y-2=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-y+4=0\\2x^2+y^2-6y-1=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2-y\\2\cdot\left(2-y\right)^2+y^2-6y-1=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=y-4\\2\cdot\left(y-4\right)^2+y^2-6y-1=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2-y\\3y^2-14y+7=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=y-4\\3y^2-22y+31=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-\dfrac{1+2\sqrt{7}}{3}\\y=\dfrac{7+2\sqrt{7}}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{-1+2\sqrt{7}}{3}\\y=\dfrac{7-2\sqrt{7}}{3}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{-1+2\sqrt{7}}{3}\\y=\dfrac{11+2\sqrt{7}}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{1+2\sqrt{7}}{3}\\y=\dfrac{11-2\sqrt{7}}{3}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

Vậy các cặp (x;y) thỏa mãn là: \(\left(-\dfrac{1+2\sqrt{7}}{3};\dfrac{7+2\sqrt{7}}{3}\right);\left(\dfrac{-1+2\sqrt{7}}{3};\dfrac{7-2\sqrt{7}}{3}\right);\left(\dfrac{-1+2\sqrt{7}}{3};\dfrac{11+2\sqrt{7}}{3}\right);\left(-\dfrac{1+2\sqrt{7}}{3};\dfrac{11-2\sqrt{7}}{3}\right)\)

6. \(\left\{{}\begin{matrix}2y-4=0\\3x+y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-2\end{matrix}\right.\)

7. \(\left\{{}\begin{matrix}4x-6y=2\\x-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\\dfrac{2+6y}{4}-\dfrac{3}{2}y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+6y}{4}\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-2\end{matrix}\right.\)

8. \(\left\{{}\begin{matrix}\dfrac{x}{3}+\dfrac{y}{2}=1\\2x+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\left(1-\dfrac{y}{2}\right).3\\6\left(1-\dfrac{y}{2}\right)+3y=\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(1-\dfrac{y}{2}\right)\\y=\left(VNghiệm\right)\end{matrix}\right.\Leftrightarrow\) không tồn tại x, y

(Các câu khác tương tự nhé.)

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

<=> \(\left\{{}\begin{matrix}3x-2y=4\\7x=14\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

2)\(\left\{{}\begin{matrix}2x+3y=5\\4x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x=6y=10\end{matrix}\right.\)

=> Hệ có vô số nghiệm.

3)\(\left\{{}\begin{matrix}3x-4y=-2\\10x+4y=28\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}3x-4y=-2\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

4)\(\left\{{}\begin{matrix}6x+15y=9\\6x-4y=28\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}6x+15y=9\\19y=19\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)