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\(x^8+x^7+1\)
\(=x^8-x^2+x^7-x+x^2+x+1\)
\(=x^2\left(x^6-1\right)+x\left(x^6-1\right)+x^2+x+1\)
\(=\left(x^2+x\right)\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x\right)\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x\right)\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^5+x^4+x^2+x\right)\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^6-x^4+x^3-x\right)\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^6-x^4+x^3-x+1\right)\left(x^2+x+1\right)\)
Chúc bạn học tốt.
x8 + x4 + 1
= x8 + 2x4 + 1 - x4
= [(x4)2 + 2x4 + 1] - x4
= (x4 + 1)2 - (x2)2
= ( x4 - x2 + 1 ) ( x4 + x2 + 1 )
x8+x+1
=(x8−x2)+(x2+x+1)
=x2(x6−1)+(x2+x+1)
=x2(x2+1)(x3−1)+(x2+x+1)
=x2(x3+1)(x−1)(x2+x+1)+(x2+x+1)
=(x2+x+1)[x2(x3+1)(x−1)+1]
=(x2+x+1)[x2(x4−x3+x−1)+1]
=(x2+x+1)(x6−x5+x3−x2+1)
Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
x8+x7+1= x8+x7+x6-x6-x5-x4+x5+x4+x3-x3-x2-x+x2+x+1
=(x8+x7+x6)-(x6+x5+x4)+(x5+x4+x3)-(x3+x2+x)+(x2+x+1)
= x6(x2+x+1)-x4(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+( x2+x+1)
=(x2+x+1)(x6-x4+x3-x+1)
Câu b, c lm tương tự
\(x^8+x^4+1\)
\(=x^4.\left(x^4+1\right)+\left(x^4+1\right)-x^4\)
\(=\left(x^4+1\right).\left(x^4+1\right)-\left(x^2\right)^2\)
\(=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+1-x^2\right).\left(x^4+1+x^2\right)\)
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
x5-x4-1=x5-x3-x2-x4+x2+x+x3-x-1
=x2.(x3-x-1)-x.(x3-x-1)+(x3-x-1)
=(x3-x-1)(x2-x+1)
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
(x+1)(x-4)(x+2)(x-8)+4x^2
=[(x+1)(x-8)][(x-4)(x+2)]+4x2
=(x2-7x-8)(x2-2x-8)+4x2
Đặt t=x2-2x-8 ta được:
(t-5x).t+4x2
=t2-5xt+4x2
=t2-xt-4xt+4x2
=t.(t-x)-4x.(t-x)
=(t-x)(t-4x)
thay t=x2-2x-8 ta được:
(x2-3x-8)(x2-6x-8)
Vậy (x+1)(x-4)(x+2)(x-8)+4x^2=(x2-3x-8)(x2-6x-8)
\(x^8+x^7+1\)
\(=x^8+x^7-x^2-x+x^2+x+1\)
\(=x^7.\left(x+1\right)-x\left(x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x+1\right)\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x.\left(x+1\right)\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x.\left(x+1\right)\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x.\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x.\left(x+1\right)\left(x-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left[x.\left(x^2-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left[\left(x^3-x\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)