\(6x^2+\left(2y-1\right)x+10y^2-28y+18=0\)

\(\Delta=\left(2y-1\right)^2-24\left(10y^2-28y+18\right)\ge0\)

\(\Leftrightarrow-236y^2+668y-431\ge0\)

\(\Rightarrow\dfrac{167-2\sqrt{615}}{118}\le y\le\dfrac{167+2\sqrt{615}}{118}\)

\(\Rightarrow y=1\)

Thế vào pt đầu ...

\(x^2-6x+9=-y^2-10y-20.\)

\(\left(x-3\right)^2=-y^2-10y-20\)

\(\left(x-3\right)^2=-y^2-10y-20\)

\(\left(x-3\right)^2=-\left(y^2+2.5y+25\right)+5\)

\(\left(x-3\right)^2=-\left(y+5\right)^2+5\)

\(\hept{\begin{cases}x=3\\y+5=\sqrt{5}\Leftrightarrow y=\sqrt{5}-5\end{cases}}\)

b)

\(\left(4x^2-4x+1\right)=-y^2-x^2-2xy\)

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

\(x=\frac{1}{2}\Leftrightarrow y=-\frac{1}{2}\)

kí hiệu a l b là a chia hết cho b nhé
 xy-1 l (x-1)(y-1) <=> xy-1 l y-1 <=> y(x-1)+y-1 l y-1 => x-1 l y-1 
tương tự : y-1 l x-1 
=> \(\orbr{\begin{cases}x-1=y-1\\x-1=1-y\end{cases}}\Rightarrow\orbr{\begin{cases}x=y\\x+y=2\end{cases}}\)

+> x=y \(\Rightarrow x^2-1\)l \(\left(x-1\right)^2\) <=> x+1 l x-1 <=> 2 l x-1 => x=2 hoặc x=3
|+> x+y=2 thay vào tương tự như trên nhé 

lm hộ t bài 1 nx

Sửa đề: \(x^2+10y^2-2xy+6y+1=0\)

\(\Leftrightarrow x^2-2xy+y^2+9y^2+6y+1=0\)

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

Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{3}\)

\(x^2+10y^2-2xy+6x+1=0\Leftrightarrow\left(x-y\right)^2+\left(3y+1\right)^2=0\)

Vì \(\left(x-y\right)^2\ge0,\left(3y+1\right)^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}x-y=0\\3y+1=0\end{matrix}\right.\)\(\Rightarrow x=y=\dfrac{-1}{3}\)