6x^2-11x-10 3x+2 2x-5 6x^2+4x - -15x-10 -15x-10 - 0

x^3-4x^2-2x+8 x-4 x^2-2 x^3-4x^2 - 0-2x+8 -2x+8 - 0

x²-x = 0 <=> x (x-1) = 0 <=> x = 0 hoặc x= 1

Với x = 0 ta có : B = 2.0⁴-11.0³+11.0²-16.0+5 = 5

Với x = 1 ta có : B = 2.1⁴-11.1³+11.1²-16.1+5 = -9

=> (x^3-2x^2+x-1)(x-5) = x^4-7x^3+11x^2-6x+5

1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

Ta có: \(2x^4+11x^3+11x^2-24x-36\)

\(=2x^4+4x^3+7x^3+14x^2-3x^2-6x-18x-36\)

\(=\left(x+2\right)\left(2x^3+7x^2-3x-18\right)\)

\(=\left(x+2\right)\left(2x^3+4x^2+3x^2+6x-9x-18\right)\)

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

\(=\left(x+2\right)^2\cdot\left(2x^2+6x-3x-9\right)\)

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

Ta có: \(x^5+7x^4+21x^3+47x^2+80x+60\)

\(=x^5+4x^4+4x^3+3x^4+12x^3+12x^2+5x^3+20x^2+20x+15x^2+60x+60\)

\(=\left(x^2+4x+4\right)\left(x^3+3x^2+5x+15\right)\)

\(=\left(x+2\right)^2\cdot\left(x+3\right)\left(x^2+5\right)\)

Ta có: \(\frac{2x^4+11x^3+11x^2-24x-36}{x^5+7x^4+21x^3+47x^2+80x+60}\)

\(=\frac{\left(x+2\right)^2\cdot\left(x+3\right)\left(2x-3\right)}{\left(x+2\right)^2\cdot\left(x+3\right)\left(x^2+5\right)}=\frac{2x-3}{x^2+5}\)

a: \(=2x^2-4x-6x+12-2x^2+10x=12\)

3x^4-17x^3+11x^2-6x+5 x-5 3x^3-2x^2+x-1 - 3x^4-15x^3 -2x^3+11x^2-6x+5 - 2x^3+10x^2 x^2-6x+5 - x^2-5x -x+5 - -x+5 0