a) \(12x^3+8x^2-3x-2=4x^2\left(3x+2\right)-\left(3x+2\right)\)

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

b)  \(18x^3+27x^2-2x-3=9x^2\left(2x+3\right)-\left(2x+3\right)\)

\(=\left(2x+3\right)\left(9x^2-1\right)=\left(2x+3\right)\left(3x-1\right)\left(3x+1\right)\)

c)  \(8x^3+4x^2-34x+15=4x^2\left(2x-3\right)+8x\left(2x-3\right)-5\left(2x-3\right)\)

\(=\left(2x-3\right)\left(4x^2+8x-5\right)=\left(2x-3\right)\left(2x-1\right)\left(2x+5\right)\)

bài 1 : ta có : \(A=27x^3+27x^2y+9xy^2+y^3=\left(3x+y\right)^3\)

\(=\left(3.\left(-3\right)+5\right)^3=\left(-9+5\right)^3=\left(-4\right)^3=-64\)

bài 2 : a) ta có : \(12a^2-3ab+8ac-2bc=3a\left(4a-b\right)+2c\left(4a-b\right)\)

\(=\left(3a+2c\right)\left(4a-b\right)\) câu này mk sữa đề lại chút .

b) ta có : câu này đề sai rồi .

nếu phân tích ra nó sẽ thành : \(17x^2+34x-5=\left(17x+17-\sqrt{374}\right)\left(x+\dfrac{17+\sqrt{374}}{17}\right)\)

c) ta có : \(4x^4+81=\left(2x^2\right)^2+36x^2+81-36x^2\)

\(=\left(2x^2+9\right)^2-36x^2=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)

câu 3 : a) ta có : \(-3x^2+2x+1=0\Leftrightarrow-3x^2+3x-x+1=0\)

\(\Leftrightarrow-3x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(-3x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\) vậy \(x=\dfrac{-1}{3};x=1\)

b) ta có : \(x\left(x-3\right)=2x-6=x\left(x-3\right)-2\left(x-3\right)=0\)

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

vậy \(x=2;x=3\)

C

\(1-27x^3\)

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

\(=\left(1-3x\right)\left(1+3x+9x^2\right)\)

\(---\)

\(x-3^3+27\)

\(=x-27+27=x\) 

\(---\)

\(27x^3+27x^2+9x+1\)

\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)

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

\(---\)

\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)

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

 \(=\left(\dfrac{x^2}{3}-y\right)^3\)

#Ayumu

1-27x\(^3\)

=(1-3x)(1+3x+9x\(^2\)

a) 8x³ - 125=(2x)³-5³=(2x-5)(2x+5)

b) 1 - 27x³y⁶=1-3³x³(y²)³=1-(3xy²)³=(1-3xy²)(1-3xy²+9x²y⁴)

c)27x³ + y³/64=(3x)³ + y³/4³=(3x)³ + (y/4)³=(3x+y/4)(9x²-3xy/4+y²/16)

d) 27x³ + 125y³ = (3x)³+(5y)³=(3x+5y)(9x²-15xy+25y²)

đề?