1.

\(x^2\)+\(y^2\)+2y-6x+10=0

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

=> (x-3)\(^2\)+(y+1)\(^2\)=0

pt vô nghiệm

4.

=> \(x^2\)+8x+16+(3y)\(^2\)-2.3.2y+4=0

=> (x+4)\(^2\)+(3y-2)\(^2\)=0

pt vô nghiệm

3.

=> (3y)\(^2\)+2.3y+1+\(x^2\)+4x+4

=> (3y+1)\(^2\)+(x+2)\(^2\)=0

pt vô nghiệm

Giúp mình với ạ,cảm ơn mọi người

b: Ta có: \(B=x^2+4x+9y^2-6y-1\)

\(=x^2+4x+4+9y^2-6y+1-6\)

\(=\left(x+2\right)^2+\left(3y-1\right)^2-6\ge-6\forall x,y\)

Dấu '=' xảy ra khi x=-2 và \(y=\dfrac{1}{3}\)

1) \(4x^2-12x+y^2-4y+13\)

\(=\left(4x^2-12x+9\right)+\left(y^2-4y+4\right)\)

\(=\left[\left(2x\right)^2-2.2x.3+3^2\right]+\left(y^2-2.2y+4\right)\)

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

2) \(x^2+y^2+2y-6x+10\)

\(=\left(x^2+2y+1\right)+\left(y^2-6x+9\right)\)

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

3) \(4x^2+9y^2-4x+6y+2\)

\(=\left(4x^2-4x+1\right)+\left(9y^2+6y+1\right)\)

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

4) \(y^2+2y+5-12x+9x^2\)

\(\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)\)

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

5) \(x^2+26+6y+9y^2-10x\)

\(=\left(x^2-10x+25\right)+\left(9y^2+6y+1\right)\)

\(=\left(x-5\right)^2+\left(3y+1\right)^2\)

Bài 1:tìm x ,biết:

a) (2x - 1)(3x + 2) - 6x(x + 1) = 0

\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)

\(\Leftrightarrow-5x=2\)

\(\Leftrightarrow x=\frac{-2}{5}\)

b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)

\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)

\(\Leftrightarrow-10x=-4\)

\(\Leftrightarrow x=\frac{2}{5}\)

c) \(4x^2-1=2\left(2x+1\right)\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)

2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)

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

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

\(=1.\left(2x-1\right)\)

c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)

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

d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)

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

e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)

\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)

\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)

Bài 1:Tìm x,y biết:

a)\(x^2-6x+y^2+10y+34\)

=>\(\left(x^2-2.x.3+3^2\right)+\left(y^2+2.y.5+5^2\right)=0\)

=>\(\left(x-3\right)^2+\left(y+5\right)^2=0\)

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

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

Còn ý b,c,d,e làm tương tự ý a.

1) \(9x^2+y^2-10y-12x+29=0\)

\(\Leftrightarrow\left(9x^2-12x+4\right)+\left(y^2-10y+25\right)=0\)

\(\Leftrightarrow\left(3x-2\right)^2+\left(y-5\right)^2=0\)

ta có : \(\left(3x-2\right)^2\ge0\forall x\) và \(\left(y-5\right)^2\ge0\forall y\)

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

vậy \(x=\dfrac{2}{3};y=5\)

2) câu này đề sai rồi nha

3) \(x^2+29+9y^2+8x-12y=0\)

\(\Leftrightarrow\left(x^2+8x+16\right)+\left(9y^2-12y+4\right)+9=0\)

\(\Leftrightarrow\left(x+4\right)^2+\left(3y-2\right)^2+9=0\)

ta có : \(\left(x+4\right)^2\ge0\forall x\) và \(\left(3y-2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x+4\right)^2+\left(3y-2\right)^2+9\ge9>0\forall x;y\)

vậy phương trình vô nghiệm

a. x² + 6x + 9 = (x + 3)²

b. 25 + 10x + x² = (5 + x)²

c. x² + 8x + 16 = (x + 4)²

d. x² + 14x + 49 = (x + 7)²

e. 4x² + 12x + 9 = (2x + 3)²

f. 9x² + 12x + 4 = (3x + 2)²

h. 16x² + 8 + 1 = (4x + 1)²

i. 4x² + 12xy + 9y² = (2x + 3y)²

k. 25x² + 20xy + 4y² = (5x + 2y)²

a) \(=\left(x+3\right)^2\)

b) \(=\left(x+5\right)^2\)

c) \(=\left(x+4\right)^2\)

d) \(=\left(x+7\right)^2\)

e) \(=\left(2x+3\right)^2\)

f) \(=\left(3x+2\right)^2\)

h) \(=\left(4x+1\right)^2\)

i) \(=\left(2x+3y\right)^2\)

k) \(=\left(5x+2y\right)^2\)