Phân tích đa thức thành nhân tử:
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
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XO
16 tháng 10 2020
(x - 5)2 - 4(x - 3)2 + 2(2x - 1)(x - 5) + (2x - 1)2
= [(x - 5)2 + 2(2x - 1)(x - 5) + (2x - 1)2) - [2(x - 3)]2
= (x - 5 + 2x - 1)2 - (2x - 6)2
= (3x - 6)2 - (2x - 6)2
= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)
16 tháng 10 2020
( x - 5 )2 - 4( x - 3 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2
= [ ( x - 5 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2 ] - 22( x - 3 )2
= ( x - 5 + 2x - 1 )2 - ( 2x - 6 )2
= ( 3x - 6 )2 - ( 2x - 6 )2
= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )
= x( 5x - 12 )
TL
2
HV
27 tháng 12 2016
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\\ =3\left(x^4-x+x^2+x+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left[\left(x^4-x\right)+\left(x^2+x+1\right)\right]-\left(x^2+x+1^2\right)\)
\(=3\left[x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2+x+1\right)\left(x^2-x+1\right)+\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(3x^2-3x+3+x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(4x^2-2x+2\right)\\ =2\left(x^2+x+1\right)\left(x^2-x+1\right)\)
27 tháng 12 2016
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)\)
\(=3\left[\left(x^4+x^3+x^2\right)-\left(x^3-x^2-x\right)+\left(x^2+x+1\right)\right]-\left(x^2+x+1\right)^2\)
\(=3\left[x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2+x+1\right)\left(x^2-x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left[3\left(x^2-x+1\right)-\left(x^2+x+1\right)\right]\)
\(=\left(x^2+x+1\right)\left(3x^2-3x+3-x^2-x-1\right)\)
\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)\)
\(=2\left(x^2+x+1\right)\left(x^2-2x+1\right)\)
\(=2\left(x^2+x+1\right)\left(x-1\right)^2\)
HL
phân tích đa thức thành nhân tử \(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
1
15 tháng 6 2017
\(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)
HD
0
NQ
2
TA
10 tháng 7 2021
`(x+3)^4+(x+5)^4-2`
`={[(x+3)^2]^2-1^2}+{[(x+5)^2]^2 -1^2}`
`=[(x+3)^2-1^2][(x+3)^2+1]+[(x+5)^2-1^2][(x+5)^2+1]`
`=(x+3-1)(x+3+1)[(x+3)^2+1]+(x+5-1)(x+5+1)[(x+5)^2+1]`
`=(x+2)(x+4)[(x+3)^2+1]+(x+4)(x+6)[(x+5)^2+1]`
`=(x+4){(x+2)[(x+3)^2+1]+(x+6)[(x+5)^2+1]}`
`=(x+4)(2x^3+24x^2+108x+176)`
10 tháng 7 2021
Bạn gì ơi hình như phải ra \(2\left(t+4\right)^2\left(x^2+8x+22\right)\)chứ nhỉ???
2 tháng 8 2018
\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(x+4+1\right)\left(x+4-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+5\right)\left(x-3\right)\)
=.= hok tốt!!
NT
0
DX
1
Ta có :
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left(x^4+x^3+x^2-x^3+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left[\left(x^4+x^3+x^2\right)-\left(x^3-1\right)\right]-\left(x^2+x+1\right)^2\)
\(=3\left[\left(x^2+x+1\right)x^2-\left(x-1\right)\left(x^2+x+1\right)\right]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2+x+1\right)\left(x^2-x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left[3\left(x^2-x+1\right)-\left(x^2+x+1\right)\right]\)
\(=\left(x^2+x+1\right)\left(3x^2-3x+3-x^2-x-1\right)\)
\(=\left(x^2+x+1\right)\left(2x^2+2-4x\right)\)
\(=2\left(x^2+x+1\right)\left(x^2+1-2x\right)\)
\(=2\left(x^2+x+1\right)\left(x-1\right)^2\)