`( 12x^6y^4  -6x^4y^3 +:3x^2y^3) : ( 3x^2y^3 )`

`=(12x^6y^4:3x^2y^3)- (6x^4y^3:3x^2y^3)+(3x^2y^3:3x^2y^3)`

`= 4x^4y - 2x^2+1`

\(a,=x\left(x^3+3x^2-6x-8\right)\\ =x\left(x^3+4x^2-x^2-4x-2x-8\right)\\ =x\left(x+4\right)\left(x^2-x-2\right)\\ =x\left(x+4\right)\left(x-2\right)\left(x+1\right)\)

\(b,=x^4+36x^2+324-36x^2\\ =\left(x^2+18\right)^2-36x^2\\ =\left(x^2+6x+18\right)\left(x^2-6x+18\right)\)

\(c,=xy\left(x^3y^3+xy+2\right)\)

a: \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)

\(=\dfrac{-6x^3y^4}{2x^3y^3}+\dfrac{4x^4y^3}{2x^3y^3}\)

\(=-3y+2x\)

b: \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}=\dfrac{5x^4y^2}{x^3y^2}-\dfrac{x^3y^2}{x^3y^2}\)

\(=5x-1\)

c: \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)

\(=-\dfrac{27x^3y^5}{3x^2y^3}-\dfrac{9x^2y^4}{3x^2y^3}+\dfrac{6x^3y^3}{3x^2y^3}\)

\(=-9xy^2-3y+2x\)

a) \(\dfrac{-6x^3y^4+4x^4y^3}{2x^3y^3}\)

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

\(=-3y+2x\)

\(=2x-3y\)

b) \(\dfrac{5x^4y^2-x^3y^2}{x^3y^2}\)

\(=\dfrac{5x\cdot x^3y^2-x^3y^2\cdot1}{x^3y^2}\)

\(=\dfrac{x^3y^2\cdot\left(5x-1\right)}{x^3y^2}\)

\(=5x-1\)

c) \(\dfrac{27x^3y^5+9x^2y^4-6x^3y^3}{-3x^2y^3}\)

\(=\dfrac{-3x^2y^3\cdot-9xy^2+-3x^2y^3\cdot-3y+-3x^2y^3\cdot2x}{-3x^2y^3}\)

\(=\dfrac{-3x^2y^3\cdot\left(-9xy^2-3y+2x\right)}{-3x^2y^3}\)

\(=-9xy^2-3x+2x\)

a, bậc 6 

b, bậc 6 

c, bậc 12 

d, bậc 9 

e, bậc 8 

a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)

b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)

c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)

\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)

d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)

hay \(N=y^2-x^2\)

a: \(=3\left(x^2-\dfrac{2}{3}x+\dfrac{4}{3}\right)\)

\(=3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{11}{9}\right)\)

\(=3\left(x-\dfrac{1}{3}\right)^2+\dfrac{11}{3}>=\dfrac{11}{3}\)

Dấu '=' xảy ra khi x=1/3

b: \(=2\left(x^2+\dfrac{3}{2}x\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}\right)\)

\(=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{9}{8}>=-\dfrac{9}{8}\)

Dấu '=' xảy ra khi x=-3/4

d: \(=3\left(x^2-2x+\dfrac{2}{3}\right)\)

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

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

Dấu '=' xảy ra khi x=1

a.\(\hept{\begin{cases}3x-2y=1\\2x+4y=3\end{cases}}\)

<=>\(\hept{\begin{cases}6x-4y=2\\2x+4y=3\end{cases}}\)

<=>\(\hept{\begin{cases}8x=5\\2x+4y=3\end{cases}}\)

<=>\(\hept{\begin{cases}x=\frac{5}{8}\\2\cdot\frac{5}{8}+4y=3\end{cases}}\)

<=>\(\hept{\begin{cases}x=\frac{5}{8}\\4y=\frac{7}{4}\end{cases}}\)

<=>\(\hept{\begin{cases}x=\frac{5}{8}\\y=\frac{7}{16}\end{cases}}\)

a) \(\hept{\begin{cases}3x-2y=1\\2x+4y=3\end{cases}}\Rightarrow\hept{\begin{cases}6x-4y=2\\2x+4y=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}8x=5\\2x+4y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{8}\\\frac{5}{4}+4y=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{5}{8}\\4y=\frac{7}{4}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{8}\\y=\frac{7}{16}\end{cases}}\)

vậy hpt có nghiệm duy nhất \(\left(x;y\right)=\left(\frac{5}{8};\frac{7}{16}\right)\)

b) \(\hept{\begin{cases}4x-3y=1\\-x+2y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}8x-6y=2\\-3x+6y=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}5x=5\\-3x+6y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\-3+6y=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\)

vậy hpt có nghiệm duy nhất \(\left(x;y\right)=\left(1;1\right)\)

a, \(\hept{\begin{cases}3x-2y=1\\2x+4y=3\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}6x-4y=2\\2x+4y=3\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}8x=5\\2x+4y=3\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{5}{8}\\4y=\frac{7}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{8}\\y=\frac{7}{16}\end{cases}}\)