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.......

Bài 1 : 

\(a)\)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+3\right)\left(x-3\right)=15\)

\(\Leftrightarrow\)\(x^3-1-x\left(x^2-3^2\right)=15\)

\(\Leftrightarrow\)\(x^3-1-x^3+9x=15\)

\(\Leftrightarrow\)\(9x=16\)

\(\Leftrightarrow\)\(x=\frac{16}{9}\)

Vậy \(x=\frac{16}{9}\)

Chúc bạn học tốt ~ 

Cái này là viết dưới dạng tổng hai bình phương ạ

a)x²-4x+5+y²+2y=x²-4x+4+y²+2y+1=(x-2)²+(y+1)²

b)2x²+y²-2xy+10x+25=x²-2xy+y²+x²+10x+25=(X+Y)²+(X+5)²

c)a²+2ab+5b²+4b+1=a²+2ab+b²+4b²+4b+1=(a+b)²+(2b+1)²

d)2x²+2b²+4x+4b+4=2x²+4x+2+2b²+4b+2=(\(\sqrt{2}x+\sqrt{2}\))²+(\(\sqrt{2}b+\sqrt{2}\))²

e)X⁴+13-6x²+4y+y²=x⁴-6x²+9+y²+4y+4=(x²-3)²+(y+2)²

f)-6x+9x²-8y+4y+y2+5= 9x²-6x+1+4y²-8y+4= (3x-1)²+(2y-2)²