1) 2x² - 4x = 2x( x - 2 )

2) 3x - 6y = 3( x - 2y )

3) x² - 3x = x( x - 3 )

4) 4x² - 6x = 2x( x - 3 )

5) x³ - 4x = x( x² - 4 ) = x( x - 2 )( x + 2 )

1) \(2x^2-4x=2x\left(x-2\right)\)

2) \(3x-6y=3\left(x-2y\right)\)

3) \(x^2-3x=x\left(x-3\right)\)

4) \(4x^2-6x=2x\left(2x-3\right)\)

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

1, \(2x^2-4x=2x\left(x-2\right)\)

2, \(3x-6y=3\left(x-2y\right)\)

3, \(x^2-3x=x\left(x-3\right)\)

4, \(4x^2-6x=2x\left(x-3\right)\)

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

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

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

(x-3)³ + 3 -x =0

=>x³-9x²+26x-24=0

=>x³-7x²+12x-2x²+14x-24=0

=>x(x²-7x+12)-2(x²-7x+12)=0

=>(x-2)(x²-7x+12)=0

=>(x-2)[x²-4x-3x+12]=0

=>(x-2)[x(x-4)-3(x-4)]=0

=>(x-2)(x-3)(x-4)=0

=>x-2=0 hoặc x-3=0 hoặc x-4=0

=>x=2 hoặc 3 hoặc 4

Vậy tập nghiệm của pt là S={2;3;4}

= (x-3)³ - (x-3) =0

(x-3)((x-3)² -1)=0

(x-3)(x-3+1)(x-3-1) =0

(x-3)(x-2)(x-4) =0

x = 3;2;4

đơn giản,dễ hiểu, vận dụng hđt đáng nhớ, có ai giỏi =em k

ai giải gimf nha

phần này học từ kì 1 rồi mà bn

co cau hoi nao dao ma bao giai

\(-\left(x+2y\right)^2\)

= \(-\left(x^2+4xy+4y^2\right)\)

= \(-\left(x+2y\right)^2\)

\(2x^2y^3-\frac{x}{4}-4y^6\)

đây là pt bậc 2 của y^3 , ta đặt y^3=z ta được

\(-\left(4z^2+\frac{2.2xz}{2}+\frac{x^2}{4}\right)+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)

\(-\left(2z+\frac{x}{2}\right)^2+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)

\(-\left\{\left(2x+\frac{x}{2}\right)^2-\left(\frac{x^2}{4}-\frac{x}{4}\right)\right\}\)

\(-\left\{\left(2x+\frac{x}{2}+\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\left(2x+\frac{x}{2}-\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\right\}\)

\(7xy^5\left(x-1\right)-3x^2y^4\left(1-x\right)+5xy^3\left(x-1\right)\)

\(=7xy^5\left(x-1\right)+3x^2y^4\left(x-1\right)+6xy^3\left(x-1\right)\)

\(=\left(x-1\right)\left(7xy^5+3x^2y^4-6xy^3\right)=xy\left(x-1\right)\left(7y^4+3xy^3-6y^2\right)\)

 Trả lời:

7xy⁵(x - 1) - 3x²y⁴(1 - x) + 5xy³(x - 1)

= 7xy⁵(x - 1) + 3x²y⁴(x - 1) + 5xy³(x - 1)

= (7xy⁵ + 3x²y⁴ + 5xy³)(x - 1)

= xy(7y⁴ + 3xy³ + 5y²)(x - 1)

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

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

x(x - y) - 2(y -x) = (x - 2)(y - x)