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a: =(x^2+x-6)(x^2+x-8)
=(x+3)(x-2)(x^2+x-8)
b: =(x^2+x)^2+4(x^2+x)-12
=(x^2+x+6)(x^2+x-2)
=(x^2+x+6)(x+2)(x-1)
c: =x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12
=(x-1)(x^3+3x^2+8x+12)
=(x-1)(x^3+2x^2+x^2+2x+6x+12)
=(x-1)(x+2)(x^2+x+6)
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a) x^2+4x+3=x^2+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3)
b) 4x^2+4x-3=4x^2+4x+1-4=(2x+1)^2-4=(2x+1-2)(2x+1+2)=(2x-1)(2x+3)
c) x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x-4)(x+3)
d) 4x^4+4x^2y^2-8y^4=4(x^4+x^2y^2-2y^4)=4(x^4-x^2y^2+2x^2y^2-2y^4)=4(x^2-y^2)(x^2+2y^2)=4(x-y)(x+y)(x^2+2y^2)
a) \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=\left(x^2+x\right)+\left(3x+3\right)\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
c) \(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
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a) x2 + 4x + 3
= x2 + 3x + x +3
= ( x2 + 3 ) + ( x + 3 )
= x ( x + 3 ) + ( x + 3 )
= ( x + 3 ) ( x + 1 )
b) 4x2 - 4x - 3
= 4x2 + 2x - 6x - 3
= ( 4x2 + 2x ) - ( 6x + 3 )
= 2x ( 2x + 1 ) - 3 ( 2x + 1 )
= ( 2x + 1 )( 2x - 3 )
c) x2 - x - 12
= x2 + 3x - 4x - 12
= ( x2 + 3x ) - ( 4x + 12 )
= x ( x + 3 ) - 4 ( x + 3 )
= ( x + 3 ) ( x - 4 )
d) 4x4 - 4x2y2 - 8y4
= 4 ( x4 - x2y2 - 2y4 )
Hk tốt
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làm khuyến mại 1 câu;
a) = 12x2 -12x2 +20x -10x +17 =0
10x = -17
x = -17/10
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a) (x2 + x)2 + 4x2 + 4x - 12
= x4 + 2x3 + x2 + 4x2 + 4x - 12
= x4 + 2x3 + 5x2 + 4x - 12
= x4 - x3 + 3x3 - 3x2 + 8x2 - 8x + 12x - 12
= x3(x - 1) + 3x2(x - 1) + 8x(x - 1) + 12(x - 1)
= (x - 1)(x3 + 3x2 + 8x + 12)
= (x - 1)(x3 + 2x2 + x2 + 2x + 6x + 12)
= (x - 1)[x2(x + 2) + x(x + 2) + 6(x + 2)]
= (x - 1)(x + 2)(x2 + x + 6)
b) (x2 + x + 1)(x2 + x + 2) - 12 (*)
Đặt x2 + x + 1 = t, ta có:
(*) = t(t + 1) - 12
= t2 + t - 12
= t2 - 3t + 4t - 12
= t(t - 3) + 4(t - 3)
= (t - 3)(t + 4)
= (x2 + x + 1 - 3)(x2 + x + 1 + 4)
= (x2 + x - 2)(x2 + x + 5)
= (x2 - x + 2x - 2)(x2 + x + 5)
= [x(x - 1) + 2(x - 1)](x2 + x + 5)
= (x - 1)(x + 2)(x2 + x + 5)
cái đề yêu cầu j z :)
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