`( 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) Kết quả bằng 3.           b) Kết quả bằng  1 2

a: \(12x^4y^3+12x^3y^3+3x^2y^3\)

\(=3x^2y^3\cdot4x^2+3x^2y^3\cdot4x+3x^2y^3\cdot1\)

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

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

b: \(x^4+xy^3-x^3y-y^4\)

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

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

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

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

a, 15x³y⁵z : 5x²y³ = 3xy²z.

b, 12x⁴y² : ( - 9xy² ) = \(\frac{3}{4}x^3\).

c, ( 30x⁴y³ - 25x²y³ - 3x⁴y⁴ ) : 5x²y³ = \(6x^2-5-\frac{3}{5}x^2y.\)

d, ( 4x⁴ - 8x²y² + 12x⁵y ) : ( - 4x² ) = -x² + 2y² - 3x³y.

a: \(=\dfrac{15}{5}\cdot\dfrac{x^3}{x^2}\cdot\dfrac{y^5}{y^3}\cdot z=3xy^2z\)

b: \(=-\dfrac{4}{3}x^3\)

c: \(=\dfrac{30x^4y^3}{5x^2y^3}-\dfrac{25x^2y^3}{5x^2y^3}-\dfrac{3x^4y^4}{5x^2y^3}\)

\(=6x^2-5-\dfrac{3}{5}x^2y\)

d: \(=\dfrac{4x^4}{-4x^2}+\dfrac{8x^2y^2}{4x^2}-\dfrac{12x^5y}{4x^2}\)

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

Câu a :

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

\(\Leftrightarrow3\left(x^2-y^2\right)+4\left(x-y\right)=0\)

\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)+4\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[3\left(x+y\right)+4\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\3\left(x+y\right)+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=-\dfrac{4}{3}\Rightarrow x=-\dfrac{4}{3}-y\end{matrix}\right.\)

Vậy \(x=y\) hoặc \(x=-\dfrac{4}{3}-y\)

Câu b :

\(\left(12x^2-3xy\right)+\left(8x-2y\right)=0\)

\(\Leftrightarrow3x\left(4x-y\right)+2\left(4x-y\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}4x-y=0\\3x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{y}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x=\dfrac{y}{4}\) hoặc \(x=-\dfrac{2}{3}\)

a, sửa đề : \(25x^2+4y^2-10x+12y+10=0\)

\(\Leftrightarrow25x^2-10x+1+4y^2+12y+9=0\)

\(\Leftrightarrow\left(5x-1\right)^2+\left(2y+3\right)^2=0\)

Đẳng thức xảy ra khi x = 1/5 ; y = -3/2 

b, \(3x^2+2y^2-12x+12y+30=0\)

\(\Leftrightarrow3\left(x^2-4x+4\right)+2\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)

Đẳng thức xảy ra khi x = 2 ; y = -3 

\(a)\)

\(25x^2+4y^2-10x+12x+10=0\)

\(\Leftrightarrow\left(5x\right)^2-10x+1+\left(2y\right)^2+12y+9=0\)

\(\Leftrightarrow[\left(5x\right)^2-10x+1+\left(2y\right)^2+12y+9=0\)

\(\Leftrightarrow[\left(5x\right)^2-2.5x.1-1^2]+[\left(2y\right)^2+2.2y.3+3^{20}]=0\)

\(\Leftrightarrow\left(5x-1\right)^2+\left(2y+3\right)^2=0\)

\(\Leftrightarrow\left(5x-1\right)^2=0\Leftrightarrow5x-1=0\Leftrightarrow x=\frac{1}{5}\)

\(\Leftrightarrow\left(2y+3\right)^2=0\Leftrightarrow2y+3=0\Leftrightarrow2y=-3\Leftrightarrow y=\frac{-3}{2}\)

\(b)\)

\(3x^2+2y^2-12x+12y+30=0\)

\(\Leftrightarrow3x^2-12x+12+2y^2+12y+18=0\)

\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)

Mà: \(3\left(x-2\right)^2\ge0\forall x;2\left(y+3\right)^2\ge0\forall y\)

\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)chỉ khi: \(x-2=y+3=0\)

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