\(2^x+12y=5y\\ \Leftrightarrow2^x=-6y\)

Với mọi \(x;y\in N\) thì \(2^x\ge1\) mà \(-6y< 0\) nên không tìm được giá trị nào của \(x;y\in N\) thỏa mãn đề bài

Vậy \(x\in\varnothing\)

\(\Leftrightarrow x^2+4y^2+9+4xy+6x+12y+y^2-1=0\)

\(\Leftrightarrow\left(x+2y+3\right)^2=1-y^2\le1\)

TH1:\(\left\{{}\begin{matrix}\left(x+2y+3\right)^2=1\\y^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-4\\x=-2\end{matrix}\right.\\y=0\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}\left(x+2y+3\right)^2=0\\y^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-5\\y=-1\Rightarrow x=-1\end{matrix}\right.\)

Vậy các cặp số nguyên t/m là \(\left(x;y\right)=\left(-4;0\right);\left(-2;0\right);\left(-5;1\right);\left(-1;-1\right)\)

H=\(x^6-2x^3+x^2-2x+2\)

\(=x^6+2x^5+3x^4+2x^2-2x^5-4x^4-6x^3-4x^2-4x+x^4+2x^3+3x^2+2x+2\)

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

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

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

\(=\left(x-1\right)^2\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]\text{≥}0\)

Vì \(\left\{{}\begin{matrix}\left(x-1\right)^2\text{≥}0\\\left(x^2+1\right)\text{≥}1\\\left(x+1\right)^2+1\text{≥}1\end{matrix}\right.\)

⇒ MinH=0 ⇔ \(x=1\)

\(x^2+5y^2+z^2+2yz-12y+2x+10=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2+2yz+z^2\right)+\left(4y^2-12y+9\right)=0\)

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

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

a) \(2x^2y^2-\frac{4}{3}x^2y+2xy\)

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

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

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

c) \(25-4x^2-y^2+4xy\)

\(=25-\left(4x^2+y^2-4xy\right)\)

\(=5^2-\left[\left(2x\right)^2-2.2x.y+y^2\right]\)

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

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

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

e) \(12y^3-3x^2y+12xy-12y\)

f) \(64x^4+y^4\)

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

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

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

a) \(2x^2y^2-\frac{4}{3}x^2y+2xy\)

\(=xy\left(2xy-\frac{4}{3}x+2\right)\)

b) 2xy².(x + 5y) - 4xy(5y + x)

= (5y + x)(2xy² - 4xy)

= 2xy(5y + x)(y - 2)

c) 25 - 4x² - y² + 4xy

= 25 - (4x² - 4xy + y²)

= 5² - (2x + y)²

= (5 - 2x - y)(5 + 2x + y)

d) x² + 4x - 2xy - 4y +y²

= (x² - 2xy + y²) + (4x - 4y)

= (x - y)² + 4(x - y)

= (x - y)(x - y + 4)

e) 12y³ - 3x²y + 12xy - 12y

= 3y(4y² - x² + 4x - 4)

= 3y[4y² - (x - 2)²]

= 3y(2y - x + 2)(2y + x - 2)

f) 64x⁴ + y⁴

= (8x²)² + 16x²y² + y⁴ - 16x²y²

= (8x² + y²)² - (4xy)²

= (8x² + y² - 4xy)(8x² + y² + 4xy)

d) x² + 4x - 2xy - 4y + y²
<=> (x^2 - 2xy + y^2) + 4(x-y)
<=> (x-y)^2 + 4(x-y)
<=> (x-y)\(\left[\left(x+y\right)+4\right]\)

Lời giải:

Ta có:

\(x^2+5y^2-4xy+6x-12y+4=0\)

\(\Leftrightarrow (x^2-4xy+4y^2)+y^2+6x-12y+4=0\)

\(\Leftrightarrow (x-2y)^2+6(x-2y)+9+y^2-5=0\)

\(\Leftrightarrow (x-2y+3)^2+y^2=5(*)\)

\(\Rightarrow y^2=5-(x-2y+3)^2\leq 5<9\)

\(\Rightarrow -3< y< 3\Rightarrow y\left\{-2;-1;0;1;2\right\}\)

Thay lần lượt các giá trị trên của $y$ vào pt $(*)$, ta thu được:

\(y=-2\Rightarrow x=-6; x=-8\)

\(y=-1\Rightarrow x=-3; x=-7\)

\(y=0\): không có $x$ thỏa mãn
\(y=1\Rightarrow x=1; x=-3\)

\(y=2\Rightarrow x=2; x=0\)