\(...P=x^2-8x+16+x^2+2xy+y^2+2y^2-2y+2\)

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

Xét \(y^2-y+1=y^2-y+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\left(\left(y-\dfrac{1}{2}\right)^2\ge0\right)\)

\(\Rightarrow2\left(y^2-y+1\right)\ge2.\dfrac{3}{4}=\dfrac{3}{2}\)

mà \(\left(x-4\right)^2\ge0;\left(x+y\right)^2\ge0\)

\(\left(1\right)\Rightarrow P\ge\dfrac{3}{2}\Rightarrow Min\left(P\right)=\dfrac{3}{2}\)

1) \(4x^5y^2-8x^4y^2+4x^3y^2\)

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

\(=4x^3y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

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

2) \(5x^4y^2-10x^3y^2+5x^2y^2\)

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

\(=5x^2y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

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

3) \(12x^2-12xy+3y^2\)

\(=3\left(4x^2-4xy+y^2\right)\)

\(=3\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

4) \(8x^3-8x^2y+2xy^2\)

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

\(=2x\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

5) \(20x^4y^2-20x^3y^3+5x^2y^4\)

\(=5x^2y^2\left(4x^2-4xy+y^2\right)\)

\(=5x^2y^2\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

1: 4x^5y^2-8x^4y^2+4x^3y^2

=4x^3y^2(x^2-2x+1)

=4x^3y^2(x-1)^2

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

3: \(=3\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)^2\)

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

5: \(=5x^2y^2\left(4x^2-4xy+y^2\right)=5x^2y^2\left(2x-y\right)^2\)

Chọn C

a: \(=\left(-\dfrac{6}{2}\right)\cdot\dfrac{x^3}{x}\cdot\dfrac{y^2}{y^2}=-3x^2\)

b: \(=\left(-\dfrac{1}{4}:\dfrac{1}{2}\right)\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^3}{y^2}=-\dfrac{1}{2}xy\)

c: \(=\dfrac{8}{4}\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^5}{y^4}=2xy\)

\(a,-6x^3y^2:2xy^2=-3x^2\)

\(b,-\dfrac{1}{4}x^4y^3:\dfrac{1}{2}x^3y^2=-\dfrac{1}{2}xy\)

\(c,8x^4y^5:4x^3y^4=2xy\)

#Urushi

\(=2y-3x+4xy-3y=-3x+4xy-y\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+8\left(x-y\right)+16=3-2y^2\)

\(\Leftrightarrow\left(x-y\right)^2+8\left(x-y\right)+16=3-2y^2\)

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

Do \(\left(x-y+4\right)^2\ge0;\forall x,y\)

\(\Rightarrow3-2y^2\ge0\Rightarrow y^2\le\dfrac{3}{2}\Rightarrow\left[{}\begin{matrix}y^2=0\\y^2=1\end{matrix}\right.\)

\(\Rightarrow y=\left\{-1;0;1\right\}\)

- Với \(y=-1\) thay vào (1):

\(\left(x+5\right)^2=1\Rightarrow\left[{}\begin{matrix}x+5=1\\x+5=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-4\\x=-6\end{matrix}\right.\)

- Với \(y=1\) thay vào (1):

\(\Rightarrow\left(x+3\right)^2=1\Rightarrow\left[{}\begin{matrix}x+3=1\\x+3=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)

- Với \(y=0\)

\(\Rightarrow\left(x+4\right)^2=3\) (ko có nghiệm nguyên do 3 ko phải SCP)