a) \(\left(5x^2-2x+1\right)\left(2x^2-3x\right)\)

\(=10x^4-15x^3-4x^3+6x^2+2x^2-3x\)

\(=10x^4-19x^3+8x^2-3x\)

b) \(\left(18x^4y^3-6x^2y^3+12x^3y^4z\right)\)

\(=6x^2y^3\left(3x^2-1+2xyz\right)\)

\(\Rightarrow\left(18x^4y^3-6x^2y^3+12x^3y^4z\right):6x^2y^3\)

\(=\left[6x^2y^3\left(3x^2-1+2xyz\right)\right]:6x^2y^3\)

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

a)(5x²-2x+1).(2x²-3x)

=10x⁴-4x³+2x²-15x³+6x²-3x

=10x⁴-19x³+8x²-3x

b)(18x⁴y³-6x²y³+12x³y⁴z):6x²y³

=(18x⁴y³:6x²y³)-(6x²y³:6x²y³)+(12x³y⁴z:6x²y³)

=3x²y-xy+2xyz

a: \(x\left(5x^2-2xy+y^2\right)=5x^3-2x^2y+xy^2\)

b: \(\left(4x-1\right)\left(2x^2-x-1\right)\)

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

\(=8x^3-6x^2-3x+1\)

\(a,=\left[x^2\left(x^2-x-1\right)+x^3+x^2-3x-1\right]:\left(x^2-x-1\right)\\ =\left[x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)+2x^2-2x-1\right]\\ =\left[x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)+2\left(x^2-x-1\right)+1\right]:\left(x^2-x-1\right)\\ =\left[\left(x^2+x+2\right)\left(x^2-x-1\right)+1\right]:\left(x^2-x-1\right)=x^2+x+2R1\)

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

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

\(=\dfrac{4x^2y\left(x^2-1\right)-3y\left(x^2-1\right)}{-\left(3x^2-2y\right)}\)

\(=\dfrac{y\left(x^2-1\right)\left(4x^2-3\right)}{-\left(3x^2-2y\right)}\)

a) `(4x^4y-7x^2y+3y).(2y-3x^2y)`

`=8x^4y^2-14x^2y^2+6y^2-12x^6y^2+21x^4y^2-9x^2y^2`

`=29x^4y^2-12x^6y^2-23x^2y^2+6y^2`

b) `(x^2+3x-3/2 x^3):2x - x/2 . (1-3/2 x)`

`=(x+3-3/2 x^2):2 - (x/2 - 3/4 x^2)`

`=x/2 + 3/2 - 3/4 x^2 -x/2 +3/4 x^2`

`=3/2`

c) `(-2x^3-x-3+5x^2):(3-2x)`

`=(3-2x)(x^2-x-1) : (3-2x)`

`=x^2-x-1`

Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)

1: Sửa đề: 3x-5

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

2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

=5x^2+14x^2+12x+8

3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

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

5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)

Vậy (2x4 – 4x3 + 5x2 + 2x – 3) : (2x2 – 1) = x2 – 2x + 3.

\(a,\left(x^3+5x^2-2x+1\right)\left(x-7\right)\\ =x^4-7x^3+5x^3-35x^2-2x^2+14x+x-7\\ =x^4-2x^3-37x^2+15x-7\\ b,\left(2x^2-3xy+y^2\right)\left(x+y\right)\\ =2x^3+2x^2y-3x^2y-3xy^2+xy^2+y^3\\ =2x^3-x^2y-2xy^2+y^3\\ c,\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\\ =x^3-5x^2+x-2x^2+10x--x^3-11x\\ =x^3-7x^2\\ d,x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)\\ =x\left(4-15x+9x^2\right)-\left(3x^2-7x-20\right)\\ =4x-15x^2+9x^3-3x^2+7x+20\\ =9x^3-18x^2+11x+20\)