Lời giải:

Ta có:

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

7x2+15x-5=0

7*(x2+2.5x-5/7)=0

x2+2*x*1.25+1,5625-255/112=0

(x+1.25)2-255/112=0

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 : \(3x^3-7x^2+17x-5\)

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

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

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

\(a,=3\left(x-5\right)-x\left(x-5\right)=\left(3-x\right)\left(x-5\right)\\ b,=7\left(x^2-2xy+y^2\right)=7\left(x-y\right)^2\\ c,=\left(x^2+y^2-2xy\right)\left(x^2+y^2+2xy\right)=\left(x-y\right)^2\left(x+y\right)^2\\ d,=\left(y^2-6y+9\right)-25x^2=\left(y-3\right)^2-25x^2=\left(y-5x-3\right)\left(y+5x-3\right)\)

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

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

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

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

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

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 !