phân tích đa thức thành nhân tử : (x+3)^4+(x+1)^4-16

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29 tháng 9 2025

Đặt \(A=\left(x+3\right)^4+\left(x+1\right)^4-16\)

\(=\left\lbrack\left(x+2\right)+1\right\rbrack^4+\left\lbrack\left(x+2\right)-1\right\rbrack^4-16\)

Đặt b=x+2

=>\(A=\left(b+1\right)^4+\left(b-1\right)^4-16\)

\(=\left(b^2+2b+1\right)^2+\left(b^2-2b+1\right)^2-16\)

\(=\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2-16\)

\(=2\left(b^2+1\right)^2+8b^2-16\)

\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-8\right\rbrack\)

\(=2\left\lbrack b^4+2b^2+1+4b^2-8\right\rbrack=2\left(b^4+6b^2-7\right)\)

\(=2\left(b^2+7\right)\left(b^2-1\right)=2\left(b^2+7\right)\left(b-1\right)\left(b+1\right)\)

\(=2\left\lbrack\left(x+2\right)^2+7\right\rbrack\left(x+2-1\right)\left(x+2+1\right)=2\left(x+1\right)\left(x+3\right)\left\lbrack\left(x+2\right)^2+7\right\rbrack\)

29 tháng 9 2025

Đặt \(A=\left(x+3\right)^4+\left(x+1\right)^4-16\)

\(=\left\lbrack\left(x+2\right)+1\right\rbrack^4+\left\lbrack\left(x+2\right)-1\right\rbrack^4-16\)

Đặt b=x+2

=>\(A=\left(b+1\right)^4+\left(b-1\right)^4-16\)

\(=\left(b^2+2b+1\right)^2+\left(b^2-2b+1\right)^2-16\)

\(=\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2-16\)

\(=2\left(b^2+1\right)^2+8b^2-16\)

\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-8\right\rbrack\)

\(=2\left\lbrack b^4+2b^2+1+4b^2-8\right\rbrack=2\left(b^4+6b^2-7\right)\)

\(=2\left(b^2+7\right)\left(b^2-1\right)=2\left(b^2+7\right)\left(b-1\right)\left(b+1\right)\)

\(=2\left\lbrack\left(x+2\right)^2+7\right\rbrack\left(x+2-1\right)\left(x+2+1\right)=2\left(x+1\right)\left(x+3\right)\left\lbrack\left(x+2\right)^2+7\right\rbrack\)

3 tháng 9 2021

nhanh hộ mình nhé.Mình đang gấp

3 tháng 7 2019

\(x^8+3x^4+4\)

\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)

\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)

\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)

3 tháng 7 2019

\(4x^4+4x^3+5x^2+2x+1\)

\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)

\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)

\(=\left(2x^2+x+1\right)^2\)

5 tháng 1 2021

1.

a. x3 - 4x2 - xy2 + 4x

= x ( x2 - 4x + 4 - y2 )

= x [ ( x - 2 )2 - y2 ]

= x ( x - y - 2 ) ( x + y - 2 )

b. x2 - x - 2 = x2 + x - 2x - 2 = x ( x + 1 ) - 2 ( x + 1 ) = ( x - 2 ) ( x + 1 )

c. x4 + 4

= ( x4 + 2x3 + 2x2 ) - ( 2x3 + 4x2 + 4x ) + ( 2x2 + 4x + 4 )

= x2 ( x2 + 2x + 2 ) - 2x ( x2 + 2x + 2 ) + 2 ( x2 + 2x + 2 )

= ( x2 + 2x + 2 ) ( x2 - 2x + 2 )

9 tháng 9 2017

x4+16= x4+ 24= (x+2)4

29 tháng 10 2021

ai giúp tớ vs

24 tháng 7 2018

\(x^4-4x^3+8x^2-16x+16 \)

\(=x^3\left(x-2\right)-2x^2\left(x-2\right)+4x\left(x-2\right)-8\left(x-2\right)\)

\(=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)

\(=\left(x-2\right)\left[x^2\left(x-2\right)+4\left(x-2\right)\right]\)

\(=\left(x-2\right)^2\left(x^2+4\right)\)

10 tháng 10 2017

\(x^{14}+16\)

\(=x^{14}+4^2\)

\(=x^{14}+8x^7+4^2-8x^7\)

\(=\left(x^7+4\right)^2-\sqrt{8x^7}^2\)

\(=\left(x^7+4-\sqrt{8x^7}\right)\left(x^7+4+\sqrt{8x^7}\right)\)

10 tháng 10 2019

\(x^{16}+4\)

\(=x^{16}-2x^{12}+2x^{12}+2x^8+2x^8-4x^8+4x^4-4x^4+4\)

\(=\left(x^{16}-2x^{12}+2x^8\right)+\left(2x^{12}-4x^8+4x^4\right)\)

\(+\left(2x^8-4x^4+4\right)\)

\(=x^8\left(x^8-2x^4+2\right)+2x^4\left(x^8-2x^4+2\right)\)

  \(+2\left(x^8-2x^4+2\right)\)

\(=\left(x^8+2x^4+2\right)\left(x^8-2x^4+2\right)\)

10 tháng 10 2019

ミ★长 - ƔξŦ★彡: Làm dài vại:(

\(x^{16}+4\)

\(=\left(x^{16}+4x^8+4\right)-4x^8\)

\(=\left(x^8+2\right)^2-\left(2x^4\right)^2\)

\(=\left(x^8+2-2x^4\right)\left(x^8+2+2x^4\right)\)

AH
Akai Haruma
Giáo viên
26 tháng 6 2024

Lời giải:

$3(x^4+x^2+1)(x^2+x+1)^2$

$=3[(x^4+2x^2+1)-x^2](x^2+x+1)^2$

$=3[(x^2+1)^2-x^2](x^2+x+1)^2$

$=3(x^2+1-x)(x^2+1+x)(x^2+x+1)^2$

$=3(x^2-x+1)(x^2+x+1)^3$