Lời giải:

Ta có:

\(3x^3-6x^2y+3xy^2-3x^2=3x(x^2-2xy+y^2-x)\)

3x³-7x²+17x-5

=3x³-x²-6x²+2x+15x-5

= x².(3x-1)-2x.(3x-1)+5.(3x-1)

= (3x-1)(x²-2x+5)

Ta có : \(3x^3-7x^2+17x-5\)

    \(=\left(3x^3-x^2\right)-\left(6x^2-2x\right)+\left(15x-5\right)\)

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

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

\(3x^3-7x^2+17x-5\)

\(=3x^3-x^2-6x^2+2x+15x-5\)

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

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

3x^3-7x^2+17x-5

= (3x^3-x^2)-(6x^2-2x)+(15x-5)

= (3x-1).(x^2-6x+15)

Tk mk nha

Ak sorry mk nhầm , kết quả là : (3x-1).(x^2-2x+5)

Tk mk nha

\(3x^3-7x^2+17x-5\)

\(=3x^3-6x^2-x^2+15x+2x-5\)

\(=\left(-6x^2+2x\right)+\left(3x^3+15x\right)-\left(x^2+5\right)\)

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

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

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

Lời giải:
$2x^2-17x+19=2(x^2-8,5x+4,25^2)-\frac{137}{8}$

$=2(x-4,25)^2-\frac{137}{8}=2[(x-\frac{17}{4})^2-\frac{137}{16}]$

$=2(x-\frac{17+\sqrt{137}}{4})(x-\frac{17-\sqrt{137}}{4})$

\(3x^3-7x^2+17x-5.\)

\(=3x^3-x^2-6x^2+2x-15x+5\)

\(=\left(3x^3-x^2\right)-\left(6x^2-2x\right)-\left(15x-5\right)\)

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

5𝑥\(^2\)−17𝑥+6

=5𝑥\(^2\)−2𝑥−15𝑥+6

=𝑥(5𝑥−2)−3(5𝑥−2)

=(𝑥−3)(5𝑥−2)

#Fiona

Chúc bạn học tốt ! 

c ) \(3x^3-7x^2+17x-5\)

\(=3x^3-x^2-6x^2+2x+15x-5\)

\(=\left(3x^3-x^2\right)-\left(6x^2-2x\right)+\left(15x-5\right)\)

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

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

câu b) )Đa thức <=> 
3.x^3 - x^2 - 6x^2 + 2x +15x - 5 
= x^2.(3x - 1) - 2x(3x-1) + 5(3x-1) 
= (3x-1).(x^2 - 2x +5) 

Tick nhé 

a> (x-2)(x^2+x+2)