Question

Tính:

+ 12x⁶y^{3 :} 4x³y

+ (x+1)(x² – x + 1)

+ 2x²y(x²+ 3xy)

Phân tích đa thức thành nhân tử:

+ 4x²y + 6 xy² -8xy

+x² – 9

+ x² – 4 +xy – 2y

+x²  - 7x +10

Tìm x biết:

+x²-x( x-2) = 2

+( x-2)² + x -2= 0

Answer 1

\(1,\\ 12x^6y^3:4x^3y=3x^3y^2\\ \left(x+1\right)\left(x^2-x+1\right)=x^3+1\\ 2x^2y\left(x^2+3xy\right)=3x^4y+6x^3y^2\\ 2,\\ a,=2xy\left(2x+3y-4\right)\\ b,=\left(x-3\right)\left(x+y\right)\\ c,=\left(x-2\right)\left(x+2\right)+y\left(x-2\right)=\left(x+y+2\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ 3,\\ a,\Leftrightarrow x^2-x^2+2x=2\\ \Leftrightarrow2x=2\Leftrightarrow x=1\\ b,\Leftrightarrow\left(x-2\right)\left(x-2+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)