a: \(\dfrac{2x^4-x^3-x^2+7x-4}{x^2+x-1}\)

\(=\dfrac{2x^4+2x^3-2x^2-3x^3-3x^2+3x+4x^2+4x-4}{x^2+x-1}\)

=2x^2-3x+4

b: \(=\dfrac{y}{x\left(2x-y\right)}+\dfrac{4x}{y\left(y-2x\right)}\)

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

c: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)

\(=\dfrac{6\left(x-2\right)^2}{x^3y}\cdot\dfrac{x^3y^2}{x-2}=6\left(x-2\right)\cdot y\)

Câu 12: B

Câu 13: C

Câu 14: A

C12 B

C13 C

C14 A

12.B     13. C         14.  A

ban chia đa thức 1 biến đã sắp xếp nha

ban chia đa thức 1 biến đã sắp xếp nha

b. (x + 2)(x² - 2x + 4)

= x³ - 2x² + 4x + 2x² - 4x + 8

= x³ + 8

c: \(\dfrac{4x^3-8x^2+13x-5}{2x-1}=\dfrac{4x^3-2x^2-6x^2+3x+10x-5}{2x-1}\)

=2x^2-3x+5

Ta có \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)

\(=\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{2x}{\left(x+2\right)\left(x-2\right)}\)

\(\frac{-4.2x}{\left(x+2\right)^2\left(x-2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{2x}=\frac{-4}{\left(x+2\right)\left(x-2\right)}\)