SORY I'M IN GRADE 6

Thợ Đào Mỏ Panda, mày bị điên à, không biết còn trả lời làm cái quái gì

Đã có đáp án:

2x^4-6x^3+x^2+6x-3

=2x^2-6x^3-3x^2-2x^2-6x-3

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

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

    \(2x^4+x^3-6x^2+x+2\) 

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

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

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

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

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

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

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

=\(\left(x+2\right)\left(x-1\right)\left(2x\left(x-1\right)+\left(x-1\right)\right)\)

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

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

a) xy+3x-7y-21

=x(y+3)-7(x+3)

=(x-7)(y+3)

b)2xy-15-6x-5y

=2x(y-3)-5(-3+y)

=(2x-5)(y-3)

c)2x^2y+2xy^2-2x-2y

=2x(xy-1)+2y(xy-1)

=(2x+2y)(xy-1)

x(x+3)-5x(x-5)-5(x+3)

=(x-5)(x+3)-5x(x-5)

=(x-5)(x+3-5x)

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

c)2x^2y+2xy^2-2x-2y=2xy(x+y)-2(x+y)=2(x+y)(xy-1)

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)

\(=x^4-x^3+3x^3-3x^2+3x^2-3x+9x-9\\ =\left(x-1\right)\left(x^3+3x^2+3x+9\right)\\ =\left(x-1\right)\left(x+3\right)\left(x^2+3\right)\)

\(x^4+2x^3+6x-9=x^3\left(x-1\right)+3x^2\left(x-1\right)+3x\left(x-1\right)+9\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+3x^2+3x+9\right)\)

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

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

\(x^4+2x^3+6x-9=0\)\(\)

\(\Rightarrow x^4+3x^3-x^3+9x-3x-9+3x^2-3x^2=0\)

\(\Rightarrow\left(x^4+3x^3+3x^2+9x\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow x\left(x^3+3x^2+3x+9\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+3\right)\left(x+3\right)=0\)

Mà \(x^2+3>0\) với mọi x\(\in\)R.

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

x³ + 2x² - 6x - 27

= (x³ - 3x²) + (5x² - 15x) + (9x - 27)

= x²(x - 3) + 5x(x - 3) + 9(x - 3)

= (x - 3)(x² + 5x + 9)   

ket ban nha

1.

= (x^3 + 125 ) -(x^2 +5x)

=(x +5) (x^2 -5x +25) -x(x+5)

=(x+5)(x^2 -5x +25 -x)

=(x+5)(x^2 -6x +25)

2.

= (x^3 -27) + (2x^2 -6x)

=(x-3) (x^2 +3x +9) +2x (x-3)

=(x-3) (x^2 +3x +9 +2x)

=(x-3) (x^2 +5x +9)

1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)

2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)

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

3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)

1) Ta có: \(x^3+2x^2-6x-27\)

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

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

2: Ta có: \(9x^2+6x-4y^2-4y\)

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

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

2x^5-6x^4-2a^2x^3-6ax^3

 =(2x^5-2a^2x^3)-(6x^4+6ax^3)

 =2x^3(x^2-a^2)-6x^3(x+a)

 =2x^3(x-a)(x+a)-6x^3(x+a)

 =(x+a)(2x^4-2x^3a-6x^3)

 =(x+a) 2x^3 (x-a-3)