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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, \(xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(x+y\right)\)
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
= \(-\left(x^2+4xy+4y^2\right)\)
= \(-\left(x+2y\right)^2\)
\(2x^2y^3-\frac{x}{4}-4y^6\)
đây là pt bậc 2 của y^3 , ta đặt y^3=z ta được
\(-\left(4z^2+\frac{2.2xz}{2}+\frac{x^2}{4}\right)+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left(2z+\frac{x}{2}\right)^2+\left(\frac{x^2}{4}-\frac{x}{4}\right)\)
\(-\left\{\left(2x+\frac{x}{2}\right)^2-\left(\frac{x^2}{4}-\frac{x}{4}\right)\right\}\)
\(-\left\{\left(2x+\frac{x}{2}+\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\left(2x+\frac{x}{2}-\sqrt{\frac{x^2}{4}-\frac{x}{4}}\right)\right\}\)
\(7xy^5\left(x-1\right)-3x^2y^4\left(1-x\right)+5xy^3\left(x-1\right)\)
\(=7xy^5\left(x-1\right)+3x^2y^4\left(x-1\right)+6xy^3\left(x-1\right)\)
\(=\left(x-1\right)\left(7xy^5+3x^2y^4-6xy^3\right)=xy\left(x-1\right)\left(7y^4+3xy^3-6y^2\right)\)
\(x\left(x-y\right)+2\left(y-x\right)=x\left(x-y\right)-2\left(x-y\right)=\left(x-y\right)\left(x-2\right)\)
\(=x\left(x-y\right)-2\left(x-y\right)=\left(x-2\right)\left(x-y\right)\)
Ta có: \(y^2\left(x^2+y\right)-zx^2-zy\)
\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)
\(=\left(x^2+y\right)\left(y^2-z\right)\)
\(x\left(x+y\right)-6x-6y\)
\(=x\left(x+y\right)-6\left(x+y\right)\)
\(=\left(x-6\right)\left(x+y\right)\)