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Tìm x
x3-x2-14x+24=0
<=> x3-3x2+2x2-6x-8x+24=0
<=> x2(x-3)+2x(x-3)-8(x-3)=0
<=> (x-3)(x2+2x-8)=0
<=> (x-3)(x2-2x+4x-8)=0
<=>(x-3)(x-2)(x+4)=0
<=> x-3=0 hay x-2=0 hay x+4=0
<=> x=3 hay x=2 hay x=-4
S={3;2;-4}
x4+7x3+14x2+14x+4
=x4+7x3+4x2+10x2+14x+4
=(x4+4x2+4)+(7x3+14x)+10x2
=(x2+2)2+7x(x2+2)+10x2
=(x2+2)2+2x(x2+2)+5x(x2+2)+10x2
=(x2+2)(x2+2+2x)+5x(x2+2+2x)
=(x2+2+2x)(x2+2+5x)
1.2x^2+x-6=2x^2+4x-3x+6=(2x^2+4x)-(3x+6)=2x(x+2)-3(x+2)=(x+2)(2x-3)
2.x^3-9x^2+14x
=x*(x^2-9x+14)
=x*(x^2-7x-2x+14)
=x*((x^2-7x)-(2x-14))
=x*(x(x-7)-2(x-7))
=x*((x--7)(x-2))
=x*(x-7)(x-2)
\(x^3-x^2-14x+24\)
\(=x^3+4x^2-5x^2-20x+6x+24\)
\(=\left(x^3+4x^2\right)-\left(5x^2+20x\right)+\left(6x+24\right)\)
\(=x^2\left(x+4\right)-5x\left(x+4\right)+6\left(x+4\right)\)
\(=\left(x^2-5x+6\right)\left(x+4\right)\)
\(=\left(x^2-2x-3x+6\right)\left(x+4\right)\)
\(=\left[x\left(x-2\right)-3\left(x-2\right)\right]\left(x+4\right)\)
\(=\left(x-2\right)\left(x-3\right)\left(x+4\right)\)
\(A=x^4-14x^3+71x^2-154x+120\)
\(=x^3\left(x-2\right)-12x^2\left(x-2\right)+47x\left(x-2\right)-60\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-12x^2+47x-60\right)\)
\(=\left(x-2\right)\left[x^2\left(x-3\right)-9x\left(x-3\right)+20\left(x-3\right)\right]\)
\(=\left(x-2\right)\left(x-3\right)\left(x^2-9x+20\right)=\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)\)
b, Vì A là tích của 4 số nguyên liên tiếp nên A chia hết cho 24
Bài 1: 4a2-4ab+b2-9a2b2
=(2a)2-2.2a.b+b2-(3ab)2
=(2a-b)2-(3ab)2
=(2a-b-3ab)(2a-b+3ab)
a/ (4a2-4ab+b2)-9a2b2
= (2a-b)2-(3ab)2
= (2a-b-3ab) (2a-b+3ab)
Mình chỉ có thể giải được 1 phần thôi:
x3 + x + 1
= x3 + 3x2 + 3x + 1 - 3x2 - 3x
= ( x3 + 3x2 + 3x + 1 ) - ( 3x2 + 3x )
= ( x + 1 )3 - 3x ( x + 1 )
= ( x + 1 ) ( x2 + 2x + 1 -3x )
= ( x + 1 ) ( x2 - x + 1 )
a) x^3 - 7x - 6
= x^3 - x - 6x - 6
= x(x^2 - 1 ) - 6 (x + 1 )
= x(x-1)(x+1) - 6 ( x + 1 )
= ( x+ 1 ) [ x(x-1) - 6 ]
= ( x + 1 )(x^2 - x - 6 )
= ( x+ 1 ) ( x^2 - 3x + 2x - 6 )
= ( x+ 1 ) [ x(x-3) + 2 ( x- 3 )]
=(x+1)(x+2)(x-3)
b) x^3 - x^2 - 14x + 24
= x^3 - 3x^2 + 2x^2 - 6x - 8x + 24
= x^2 ( x - 3 ) + 2x(x-3) - 8 ( x- 3 )
= ( x - 3 )( x^2 + 2 x - 8 )
= ( x- 3 ) [ x^2 + 4x - 2x - 8 )]
= ( x- 3 )( [ x( x + 4 ) - 2 ( x+ 4) ]
= ( x - 3 )( x+ 4 )( x- 2 )
c) x^5 + x + 1
= x^5 - x^2 + x^2 + x + 1
= x^2(x^3 - 1 ) + x^2 + x + 1
= x^2 ( x- 1 )(x^2 + x + 1 ) + x^2 + x+ 1
= ( x^2 + x + 1 )( x^3 - x^2 ) + x^2 + x + 1
=( x^2 + x + 1 )( X^3 - x^2 + 1 )
x2+7x+6
giúp mình nha (phân tích thành nhân tử )
nhanh mình k cho
x4+2.x3-13.x2-14x+24
=x3.(x+2)-13x2+12x-26x+24
=x3.(x+2)-x.(13x-12)-2.(13x-12)
=x3.(x+2)-(13x-12)(x+2)
=(x+2)(x3-13x+12)
=(x+2)(x3-x-12x+12)
=(x+2)[x.(x2-1)-12.(x-1)]
=(x+2)[x.(x-1)(x+1)-12.(x-1)]
=(x+2)(x-1)[x.(x+1)-12]
=(x+2)(x-1)(x2+x-12)
=(x+2)(x-1)(x2-3x+4x-12)
=(x+2)(x-1)[x.(x-3)+4.(x-3)]
=(x+2)(x-1)(x-3)(x+4)