\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x-3\right)^2\)

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

2x³- 12x² + 18x = 2x (x² - 6x + 9 ) = 2x ( x - 3 )²

1, \(xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(x+y\right)\)

2, \(5x\left(3y+4x-6\right)\)

3, \(3x\left(2-y\right)\)

4, \(x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)

5, \(x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\)

6, \(2xy\left(x+2y-5x^2y\right)\)

7, \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

11, \(\left(x+y\right)\left(x-1\right)\)

\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)

:) bóc lột !

Câu 1: 

a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x

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

\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)

\(=2x^2+6x+17\)

c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)

Câu 2: 

\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x^2-2.x.3+3^2\right)=2x\left(x-3\right)^2\)

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

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

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

5𝑥\(^2\)−17𝑥+6

=5𝑥\(^2\)−2𝑥−15𝑥+6

=𝑥(5𝑥−2)−3(5𝑥−2)

=(𝑥−3)(5𝑥−2)

#Fiona

Chúc bạn học tốt ! 

1. 

\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)

2. 

\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt

Bài 1:

Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)

\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)

\(=6x\left(2x+1\right)\cdot2y\)

\(=12xy\left(2x+1\right)\)

Bài 2: 

Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)

a: x^3-2x-4=0

=>x^3-2x^2+2x^2-4x+2x-4=0

=>(x-2)(x^2+2x+2)=0

=>x-2=0

=>x=2

b: 2x^3-12x^2+17x-2=0

=>2x^3-4x^2-8x^2+16x+x-2=0

=>(x-2)(2x^2-4x+1)=0

=>x=2; \(x=\dfrac{4\pm\sqrt{14}}{2}\)

\(=-12x^2\left(y-x\right)+18x^3\left(y-x\right)\)

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

anh cho em hỏi anh tử là j ạ

đây là đa thức bậc 4 không phải phân thức

ý em bảo là nhân tự ạ