từ phương trình số 2 ta có 
\(\left(x+y\right)\left(x+2y\right)+\left(x+y\right)=0\Leftrightarrow\left(x+y\right)\left(x+2y+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+y=0\\x+2y+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-y\\x=-2y-1\end{cases}}\)

lần lượt thay vào 1 ta có 

\(\orbr{\begin{cases}y^2+7=y^2+4y\\\left(-2y-1\right)^2+7=y^2+4y\end{cases}\Leftrightarrow\orbr{\begin{cases}y=\frac{7}{4}\\3y^2+8=0\end{cases}}}\)

vậy hệ có nghiệm duy nhất \(x=-y=-\frac{7}{4}\)

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2

\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

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

a,9x^2+y^2+2z^2−18x+4z−6y+20=0

⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0

⇔x=1;y=3;z=−1

b,5x^2+5y^2+8xy+2y−2x+2=0

⇔4(x+y)2+(x−1)2+(y+1)2=0

⇔x=−y;x=1y=−1⇔x=1y=−1

c,5x^2+2y^2+4xy−2x+4y+5=0

⇔(2x+y)^2+(x−1)^2+(y+2)^2=0

⇔2x=−y;x=1;y=−2

⇔x=1;y=−2

⇔(x−1)^2+(2y−3)^2+(z+2)^2=0

\(d,\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

\(\Rightarrow\)PT vô nghiệm vì 11 không phải là tổng 2 số chính phương

Ta có:

\(\left\{{}\begin{matrix}x^2+2y^2-3xy-2x+4y=0\\\left(x^2-5\right)^2=2x-2y+5\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x-2y\right)=0\\x^4-10x^2+25=2x-2y+5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-y-2=0\\x-2y=0\end{matrix}\right.\\x^4-10x^2+20-2x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x-2\\x^4-10x^2+20-2x+2\left(x-2\right)=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\x^4-10x^2+20-2x+\dfrac{2x}{2}=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x-2\\x^4-10x^2+16=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\x^4-10x^2-x+20=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x-2\\\left(x^2-8\right)\left(x^2-2\right)=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\\left(x^2-x-5\right)\left(x^2+x-4\right)=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x-2\\\left[{}\begin{matrix}x^2=8\\x^2=2\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\\left[{}\begin{matrix}x^2-x-5=0\\x^2+x-4=0\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=x-2\\\left[{}\begin{matrix}x=\sqrt{8}\\x=-\sqrt{8}\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}y=x-2\\\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\\left[{}\begin{matrix}x=\dfrac{1+\sqrt{21}}{2}\\x=\dfrac{1-\sqrt{21}}{2}\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}y=\dfrac{x}{2}\\\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{17}}{2}\\x=\dfrac{-1-\sqrt{17}}{2}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}y=\sqrt{8}-2\\x=\sqrt{8}\end{matrix}\right.\\\left\{{}\begin{matrix}y=-\sqrt{8}-2\\x=-\sqrt{8}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}\left\{{}\begin{matrix}y=\sqrt{2}-2\\x=\sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}y=-\sqrt{2}-2\\x=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}y=\dfrac{1+\sqrt{21}}{4}\\x=\dfrac{1+\sqrt{21}}{2}\end{matrix}\right.\\\end{matrix}\right.\) (CÒN MỘT VÀI TRƯỜNG HỢP BÊN TRÊN MK KO VIẾT HẾT ĐƯỢC BẠN TỰ TÌM Y NHA)

a) Quy đồng mẫu thức và sử dụng hằng đẳng thức rồi rút gọn thu được x + 1 2 ( x − 1 )  

b) Tương tự a) thu được 2 2 − y

$A=x^2+y^2-6x+4y+20=(x^2-6x+9)+(y^2+4y+4)+7$

$=(x-3)^2+(y+2)^2+7\geq 0+0+7=7$
Vậy $A_{\min}=7$. Giá trị này đạt tại $(x-3)^2=(y+2)^2=0$

$\Leftrightarrow x=3; y=-2$

---------------------

$B=9x^2+y^2+2z^2-18x+4z-6y+30$

$=(9x^2-18x+9)+(y^2-6y+9)+(2z^2+4z+2)+10$

$=9(x^2-2x+1)+(y^2-6y+9)+2(z^2+2z+1)+10$

$=9(x-1)^2+(y-3)^2+2(z+1)^2+10\geq 10$
Vậy $B_{\min}=10$. Giá trị này đạt tại $(x-1)^2=(y-3)^2=(z+1)^2$

$\Leftrightarrow x=1; y=3; z=-1$

$C=x^2+y^2+z^2-xy-yz-xz+3$

$2C=2x^2+2y^2+2z^2-2xy-2yz-2xz+6$

$=(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)+6$

$=(x-y)^2+(y-z)^2+(z-x)^2+6\geq 6$

$\Rightarrow C\geq 3$

Vậy $C_{\min}=3$. Giá trị này đạt tại $x-y=y-z=z-x=0$

$\Leftrihgtarrow x=y=z$

--------------------------------------

$D=5x^2+2y^2+4xy-2x+4y+2021$

$=2(y^2+2xy+x^2)+3x^2-2x+4y+2021$

$=2(x+y)^2+4(x+y)+3x^2-6x+2021$
$=2(x+y)^2+4(x+y)+2+3(x^2-2x+1)+2016$

$=2[(x+y)^2+2(x+y)+1]+3(x^2-2x+1)+2016$

$=2(x+y+1)^2+3(x-1)^2+2016\geq 2016$

Vậy $D_{\min}=2016$ khi $x+y+1=x-1=0$

$\Leftrightarrow x=1; y=-2$

$E=x^2-2x+4y^2+4y+2014$

$=(x^2-2x+1)+(4y^2+4y+1)+2012$

$=(x-1)^2+(2y+1)^2+2012$

$\geq 2012$

Vậy $E_{\min}=2012$. Giá trị này đạt tại $x-1=2y+1=0$

$\Leftrightarrow x=1; y=\frac{-1}{2}$

----------------------

$F=5x^2+5y^2+8xy+2y-2x+30$

$=4(x^2+2xy+y^2)+x^2+y^2+2y-2x+30$

$=4(x+y)^2+(x^2-2x+1)+(y^2+2y+1)+28$

$=4(x+y)^2+(x-1)^2+(y+1)^2+28\geq 28$

Vậy $F_{\min}=28$. Giá trị này đạt tại $x+y=x-1=y+1=0$

$\Leftrightarrow x=1; y=-1$

Xét \(x^2-3xy+y^2=0\)

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

Đơn giản rồi nhé