\(4x^2-4x+1+9y^2-6y+1=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\3y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Ta có:4x²-4x+9y²-6y+2=0

   <=>(4x²-4x+1)+(9y²-6y+1)=0

   <=> (2x-1)²+(3y-1)²=0

   \(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\3y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

b) Ta có: \(B=x^2+2x+y^2-4y+6\)

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

\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)

c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)

\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)

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

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

\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2

\(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)

dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)

\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)

=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)

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

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

dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

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

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

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

\(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]=\left(3x-3y-2x+2y\right)\left(3x-3y+2x+2y\right)=\left(x-y\right)\left(5x-y\right)\)

\(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

\(27x^3-0,001=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)

\(125x^3-1=\left(5x-1\right)\left(25x^2+5x+1\right)\)

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

c: \(9\left(x-y\right)^2-4\left(x+y\right)^2=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)=\left(x-5y\right)\left(5x-y\right)\)

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

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

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

e: \(x^3+27=\left(x+3\right)\left(x^2+3x+9\right)\)

x²+y²-4x+6y+13=0

(x²-4x+4)+(y²+6y+9)=0

(x-2)²+(y+3)²=0

suy ra x-2=0 hoặc y+3=0

*x-2=0=>x=2      *y+3 =0=> y=-3

vậy x=2,y=-3

x² + 2y² + 2xy - 4x + 6y + 29 = 0

<=> ( x² + 2xy + y² - 4x - 4y + 4 ) + ( y² + 10y + 25 ) = 0

<=> [ ( x² + 2xy + y² ) - 2( x + y ).2 + 2² ] + ( y + 5 )² = 0

<=> ( x + y - 2 )² + ( y + 5 )² = 0 (*)

<=> \(\hept{\begin{cases}\left(x+y-2\right)^2\ge0\forall x,y\\\left(y+5\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x+y-2\right)^2+\left(y+5\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}x+y-2=0\\y+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=7\\y=-5\end{cases}}\)

Vậy x = 7 ; y = -5