phân tích đa thức thành nhân tử
\(x^{16}\)+4
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KD
2
NL
Nguyễn Lê Phước Thịnh
CTVHS
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
=>\(A=\left(b+1\right)^4+\left(b-1\right)^4-16\)
\(=\left(b^2+2b+1\right)^2+\left(b^2-2b+1\right)^2-16\)
\(=\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2-16\)
\(=2\left(b^2+1\right)^2+8b^2-16\)
\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-8\right\rbrack\)
\(=2\left\lbrack b^4+2b^2+1+4b^2-8\right\rbrack=2\left(b^4+6b^2-7\right)\)
\(=2\left(b^2+7\right)\left(b^2-1\right)=2\left(b^2+7\right)\left(b-1\right)\left(b+1\right)\)
\(=2\left\lbrack\left(x+2\right)^2+7\right\rbrack\left(x+2-1\right)\left(x+2+1\right)=2\left(x+1\right)\left(x+3\right)\left\lbrack\left(x+2\right)^2+7\right\rbrack\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
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
=>\(A=\left(b+1\right)^4+\left(b-1\right)^4-16\)
\(=\left(b^2+2b+1\right)^2+\left(b^2-2b+1\right)^2-16\)
\(=\left(b^2+1\right)^2+4b\left(b^2+1\right)+4b^2+\left(b^2+1\right)^2-4b\left(b^2+1\right)+4b^2-16\)
\(=2\left(b^2+1\right)^2+8b^2-16\)
\(=2\left\lbrack\left(b^2+1\right)^2+4b^2-8\right\rbrack\)
\(=2\left\lbrack b^4+2b^2+1+4b^2-8\right\rbrack=2\left(b^4+6b^2-7\right)\)
\(=2\left(b^2+7\right)\left(b^2-1\right)=2\left(b^2+7\right)\left(b-1\right)\left(b+1\right)\)
\(=2\left\lbrack\left(x+2\right)^2+7\right\rbrack\left(x+2-1\right)\left(x+2+1\right)=2\left(x+1\right)\left(x+3\right)\left\lbrack\left(x+2\right)^2+7\right\rbrack\)
MT
1
P
10 tháng 6 2023
\(x^4+3x^3+12x-16\)
\(=x^4+4x^3+4x^2+16x-x^3-4x^2-4x-16\)
\(=x\left(x^3+4x^2+4x+16\right)-\left(x^3+4x^2+4x+16\right)\)
\(=\left(x-1\right)\left(x^3+4x^2+4x+16\right)\)
\(=\left(x-1\right)\left[x^2\left(x+4\right)+4\left(x+4\right)\right]\)
\(=\left(x-1\right)\left(x+4\right)\left(x^2+4\right)\)
QA
1
VA
1
AH
13 tháng 8 2018
mình biết nội quy rồi nên đưng đăng nội quy
ai chơi bang bang 2 kết bạn với mình
mình có nick có 54k vàng đang góp mua pika
ai kết bạn mình cho
TT
1
R
0
LL
0
HN
1
HP
1 tháng 8 2016
Ta có:
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16\right)+8\left(x^2+10x+16\right)+16\)
\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)
\(=\left(x^2+10x+16+4\right)^2=\left(x^2+10+20\right)^2\)
k nha!!
6 tháng 10 2016
\(\text{( x + 2 ) ( x + 4 ) ( x + 6 ) ( x + 8 ) + 16}\)
\(\text{Phân tích thành nhân tử :}\)
\(\left(x^2+10x+20\right)^2\)
\(x^{16}+4\)
\(=x^{16}-2x^{12}+2x^{12}+2x^8+2x^8-4x^8+4x^4-4x^4+4\)
\(=\left(x^{16}-2x^{12}+2x^8\right)+\left(2x^{12}-4x^8+4x^4\right)\)
\(+\left(2x^8-4x^4+4\right)\)
\(=x^8\left(x^8-2x^4+2\right)+2x^4\left(x^8-2x^4+2\right)\)
\(+2\left(x^8-2x^4+2\right)\)
\(=\left(x^8+2x^4+2\right)\left(x^8-2x^4+2\right)\)
ミ★长 - ƔξŦ★彡: Làm dài vại:(
\(x^{16}+4\)
\(=\left(x^{16}+4x^8+4\right)-4x^8\)
\(=\left(x^8+2\right)^2-\left(2x^4\right)^2\)
\(=\left(x^8+2-2x^4\right)\left(x^8+2+2x^4\right)\)