Phương trình tương đương (3x)2+2.3x+1+(2y)2−2.2x.2+4=0(3x)2+2.3x+1+(2y)2−2.2x.2+4=0 ⇒(3x+1)2+(2y−2)2=0⇒(3x+1)2+(2y−2)2=0 Do (3x+1)2≥0(3x+1)2≥0 và (2y−2)2≥0(2y−2)2≥0 ∀x,y∀x,y ⇒(3x+1)2+(2y−2)2≥0⇒(3x+1)2+(2y−2)2≥0 Dấu "=" xảy ra ⇔⇔ ⇒{(3x+1)2=0(2y−2)2=0⇒{(3x+1)2=0(2y−2)2=0 ⇒{3x+1=02y−2=0⇒{3x+1=02y−2=0 ⇒⎧⎨⎩x=−13y=1
hok tốt
\(9x^2+6x+4y^2-8y+5=0\)
\(\Leftrightarrow9x^2+6x+1+4\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)^2+4\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}3x+1=0\\y-1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{3}\\y=1\end{cases}}\)
vậy.......
\(9x^2+6x+4y^2-8y+5=0\)
\(\Leftrightarrow\left(9x^2+6x+1\right)+\left(4y^2-8y+4\right)=0\)
\(\Leftrightarrow\left(3x+1\right)^2+4\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)^2+4\left(y-1\right)^2=0\)(1)
Vì \(\left(3x+1\right)^2\ge0\forall x\); \(4\left(y-1\right)^2\ge0\forall y\)
\(\Rightarrow\left(3x+1\right)^2+4\left(y-1\right)^2\ge0\forall x,y\)(2)
Từ (1) và (2) \(\Rightarrow\left(3x+1\right)^2+4\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}3x+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=-1\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{3}\\y=1\end{cases}}\)
Vậy \(x=-\frac{1}{3}\); \(y=1\)
