Ta có: \(\left\{{}\begin{matrix}x^4+2x^3y+x^2y^2=2x+9\\x^2+2xy=6x+6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+xy\right)^2=2x+9\\x^2+2xy=6x+6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+xy\right)^2=2x+9\\xy=3x+3-\dfrac{x^2}{2}\end{matrix}\right.\) \(\Rightarrow\left(\dfrac{x^2}{2}+3x+3\right)^2=2x+9\)( đến đây là phương trình 1 ẩn rồi, tự giải tiếp)

1. Đề này là 18 chứ không phải 15 nhé

\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+x+\sqrt{y^2+x+y+1}+y=18\left(1\right)\\\sqrt{x^2+x+y+1}-x+\sqrt{y^2+x+y+1}-y=2\left(2\right)\end{matrix}\right.\)

Lấy (1) + (2) và (1) - (2) ta được hệ mới

\(\left\{{}\begin{matrix}\sqrt{x^2+x+y+1}+\sqrt{y^2+x+y+1}=10\\x+y=8\end{matrix}\right.\)

\(\Rightarrow x=8-y\)

\(\Rightarrow\sqrt{x^2+9}+\sqrt{y^2+9}=10\)\(\Leftrightarrow\sqrt{x^2+9}=10-\sqrt{y^2+9}\)

\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\x^2+9=100-20\sqrt{y^2+9}+y^2+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\x^2=100-20\sqrt{y^2+9}+y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\\left(8-y\right)^2=100-20\sqrt{y^2+9}+y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10-\sqrt{y^2+9}>0\\9y^2-72y+144=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)

2. Dễ thấy x = y = 0 không phải là nghiệm của phương trình

HPT\(\Leftrightarrow\left\{{}\begin{matrix}1-\dfrac{12}{y+3x}=\dfrac{2}{\sqrt{x}}\left(1\right)\\1+\dfrac{12}{y+3x}=\dfrac{6}{\sqrt{y}}\left(2\right)\end{matrix}\right.\)

Lấy (1) + (2) ; (1) - (2) ta được

\(\left\{{}\begin{matrix}1=\dfrac{1}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}\left(3\right)\\\dfrac{12}{y+3x}=\dfrac{3}{\sqrt{y}}-\dfrac{1}{\sqrt{x}}\left(4\right)\end{matrix}\right.\)

Lấy ( 3) nhân (4)

\(\dfrac{12}{y+3x}=\dfrac{9}{y}-\dfrac{1}{x}=\dfrac{9x-y}{xy}\)

\(\Leftrightarrow27x^2-6xy-y^2=0\Leftrightarrow\left(9x+y\right)\left(3x-y\right)=0\)

\(\Rightarrow y=3x\)

đến đây thì dễ rồi

3. Đây là hệ đối xứng loại I

\(\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=8\\\left(x+y\right)+2xy=2\end{matrix}\right.\)

Đặt S = a + b ; P = ab (\(S^2\ge4P\) )

xong giải ra thôi mà

1)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(-1;2\right)\)

2)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

3)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;1\right)\)

4) 

HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

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.

a.

\(x^2-3y^2+2xy-x+5y-2=0\)

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

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

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

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

Thay lên pt đầu: \(\left[{}\begin{matrix}\left(y-1\right)^2+y^2+y-1+y=8\\\left(2-3y\right)^2+y^2+2-3y+y=8\end{matrix}\right.\)

Bạn tự giải nốt

b.

\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=9-2xy\\4x+6y=20-2xy\end{matrix}\right.\)

\(\Rightarrow x+y=11\Rightarrow y=11-x\)

Thay vào pt đầu:

\(3x+5\left(11-x\right)=9-2x\left(11-x\right)\)

Bạn tự giải nốt

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:

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

\(\Rightarrow x=4\)

e.

\(\left\{{}\begin{matrix}2x-3y+5=0\\3x+5y-21=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y=-25\\9x+15y=63\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=38\\3x+5y=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{21-3x}{5}\end{matrix}\right.\)

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

f.

\(\left\{{}\begin{matrix}x-y\sqrt{2}=0\\2x\sqrt{2}+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y\sqrt{2}=0\\4x+y\sqrt{2}=5\sqrt{2}\end{matrix}\right.\)

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

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

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

a.

\(\Leftrightarrow\left\{{}\begin{matrix}5x=-25\\3x-5y=-30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{3x+30}{5}\end{matrix}\right.\)

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

b.

\(\Leftrightarrow\left\{{}\begin{matrix}8x-6y=-10\\9x+6y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}17x=-34\\9x+6y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=\dfrac{-24-9x}{6}\end{matrix}\right.\)

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

c.

\(\left\{{}\begin{matrix}3x+3y=9\\4x-2y=-2\end{matrix}\right.\)

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

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

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

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

d.

\(\left\{{}\begin{matrix}5x-4y=32\\6x+2y=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-4y=32\\12x+4y=36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-4y=32\\17x=68\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{3x-32}{4}\end{matrix}\right.\)

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

\(\left\{{}\begin{matrix}x^2-2xy-3y^2=0\\x^2+y^2+2x+3y=19\end{matrix}\right.\) giải PT \(x^2-2xy-3y^2=0\)

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

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

+) Nếu x=-y Thay vào PT thứ 2 ta được

\(\left(-y\right)^2+y^2+2\left(-y\right)+3y=19\) \(\Leftrightarrow2y^2+y=19\) \(\Leftrightarrow2y^2+y-19=0\)

Đây là PT bậc 2 ẩn y nên\(\Delta=1^2+2.4.19=153>0\)

\(\Rightarrow\) PT có 2 nghiệm phân biệt \(\left[{}\begin{matrix}y=\frac{-1+3\sqrt{17}}{4}\Rightarrow x=\frac{1-3\sqrt{17}}{4}\\y=\frac{-1-3\sqrt{17}}{4}\Rightarrow x=\frac{1+3\sqrt{17}}{4}\end{matrix}\right.\) (thỏa mãn)

+) Nếu x=3y Thay vào PT thứ 2 ta được

\(\left(3y\right)^2+y^2+2.3y+3y=19\) \(\Leftrightarrow9y^2+y^2+6y+3y=19\)

\(\Leftrightarrow10y^2+9y=19\) \(\Leftrightarrow10y^2+9y-19=0\) \(\Leftrightarrow\left(10y^2-10y\right)+\left(19y-19\right)=0\) \(\Leftrightarrow10y\left(y-1\right)+19\left(y-1\right)=0\)

\(\Leftrightarrow\left(10y+19\right)\left(y-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}y=1\Rightarrow x=3\\y=\frac{-19}{10}\Rightarrow x=\frac{-57}{10}\end{matrix}\right.\) (thỏa mãn)

Vậy HPT có 4 cặp nghiệm (x,y)là\(\left(\frac{1-3\sqrt{17}}{4};\frac{-1+3\sqrt{17}}{4}\right);\)\(\left(\frac{1+3\sqrt{17}}{4};\frac{-1-3\sqrt{17}}{4}\right)\);(3;1);

\(\left(\frac{-57}{10};\frac{-19}{10}\right)\)