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\(4x^4-8x^3+3x^2-8x+4\)
\(=\left(4x^4-8x^3\right)+\left(3x^2-6x\right)-\left(2x-4\right)\)
\(=4x^3\left(x-2\right)+3x\left(x-2\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(4x^3+3x-2\right)\)
x4 - 4x3 - 8x2 + 8x
= x(x3 - 4x2 - 8x + 8)
= x[x3 + 8 - 4x(x + 2)]
= x[(x + 2)(x2 - 2x + 4) - 4x(x + 2)]
= x(x + 2)(x2 - 6x + 4)
= x(x + 2)(x2 - 6x + 9 - 5)
= \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x^3-6x^2+2x^2+4x-12x+8\right)\)
\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)
\(=x\left[x\left(x^2-6x+4\right)+2\left(x^2-6x+4\right)\right]\)
\(=x\left(x^2-6x+4\right)\left(x+2\right)\)
\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)
\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)
\(=\left(2x\right)^3-\left(4y^2\right)^3\)
Sau đó thì sử dụng HĐT số 7
8x3 + 27x3 + 4x2 + 9y2 - 6xy
= (2x + 3y)(4x2 - 6xy + 9y2) + (4x2 - 6xy +9y2)
= (4x2 - 6xy + 9y2)(2x + 3y + 1)
Không chắc lắm
a) -8x2+5x+3=-8x^2+8x-3x+3=-8x(x-1)-3(x-1)=-(8x+3)(x-1)
b)8x^2-10x-3=8x^2-12x+2x-3=8x(x-1,5)+2(x-1,5)=2(4x+1)(x-1,5)
c)=8x^2-2x+12x-3=2x(4x-1)+3(4x-1)=(2x+3)(4x-1)
d)=-8x^2+24x-x+3=-8x(x-3)-(x-3)=-(8x+1)(x-3)
x3 - 2x2 - 8x
= x( x2 - 2x - 8 )
= x( x2 - 4x + 2x - 8 )
= x[ x( x - 4 ) + 2( x - 4 ) ]
= x( x - 4 )( x + 2 )
\(x^3-2x^2-8x=x\left(x^2-2x-8\right)=x\left(x^2-2x+1-9\right)=x\left[\left(x-1\right)^2-3^2\right]=x\left(x-4\right)\left(x+2\right)\)
8x2 - 2x - 3 = 8x2 + 4x - 6x - 3
= 4x( 2x + 1 ) - 3( 2x + 1)
= ( 2x + 1 )( 4x - 3 )
4x2 - 8x + 3
= 4x2 - 6x - 2x + 3
= ( 4x2 - 6x ) - ( 2x - 3 )
= 2x( 2x - 3 ) - ( 2x - 3 )
= ( 2x - 3 )( 2x - 1 )
\(4x^2-8x+3\)
\(=4x^2-2x-6x+3\)
\(=\left(4x^2-6x\right)-\left(2x-3\right)\)
\(=2x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(2x-1\right)\left(2x-3\right)\)