Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^8-1=x^8-x^4+x^4-1\)
\(=x^4\left(x^2-1\right)+\left(x^4-1\right)\)
\(=x^4\left(x^2-1\right)+\left(x^2-1\right)\left(x^2+1\right)\)
\(=\left(x^2-1\right)\left(x^4+x^2+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)\)
\(x^8-1\)
\(=\left(x^4\right)^2-1^2\)
\(=\left(x^4-1\right)\left(x^4+1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\)
#) TL :
x8 + x4 + 1
= (x4)2 + 2x4 + 1 - x4
= ( x4 + 1 )2 - x4
= ( x4 - x2 + 1 )(x4 + x2 + 1)
= ( x4 - x2 + 1)( x2 - x + 1)( x2 + x + 1 )
Chúc bn hok tốt ạ :3
= (4x^4+4x^2+1)-4x^2
=(2x^2+1)^2-(2x)^2
= (2x^2-2x+1)(2x^2+2x+1)
\(a^6+a^4+a^2b^2+b^4-b^6\)
\(=(a^2)^3-(b^2)^3+(a^4+a^2b^2+b^4)\)
\(=(a^2-b^2)(a^4+a^2b^2+b^4)+(a^4+a^2b^2+b^4)\)
\(=(a^2-b^2+1)(a^4+a^2b^2+b^4)\)
\(=(a^4+2a^2b^2+b^4-a^2b^2)(a^2-b^2+1)\)
\(=(a^2+ab+b^2)(a^2-ab+b^2)(a^2-b^2+1)\)
\(a^6+a^2b^2+a^4+b^2-b^6\)
\(=a^4\left(a^2+b^2\right)+a^2\left(a^2+b^2\right)-b^6\)
\(=\left(a^2+b^2\right)+\left(a^4+a^2\right)-b^6\)
x\(^2\)-(a+b)x+ab
= x\(^2\)-ax-bx+ab
= x(x-a) - b(x-a)
= ( x-a).( x-b)
ax-2x-a\(^2\)+2a
= x(a-2) - a(a-2)
= (a-2).( x-a)
\(a^8+a^4+1=\left(a^8+2a^4+1\right)-a^4\)
\(=\left(a^4+1\right)^2-a^4\)
\(=\left(a^4-a^2+1\right)\left(a^4+a^2+1\right)\)
\(=\left[\left(a^4-2a^2+1\right)-a^2\right]\left(a^4+a^2+1\right)\)
\(=\left[\left(a^2-1\right)^2-a^2\right]\left(a^4+a^2+1\right)\)
\(=\left(a^2-a-1\right)\left(a^2-a+1\right)\left(a^4+a^2+1\right)\)
*\(a^8+a^7+1=a^8+a^7+a^6-a^6+a^5-a^5+a^4-a^4+a^3-a^3+a^2-a^2+a-a+1\)\(=\left(a^8+a^7+a^6\right)+\left(a^5+a^4+a^3\right)+\left(a^2+a+1\right)-\left(a^6+a^5+a^4\right)-\left(a^3+a^2+a\right)\)\(=a^6\left(a^2+a+1\right)+a^3\left(a^2+a+1\right)+\left(a^2+a+1\right)-a^4\left(a^2+a+1\right)-a\left(a^2+a+1\right)\)\(=\left(a^2+a+1\right)\left(a^6-a^4+a^3-a+1\right)\)