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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
Ta có :
x7 + x5 + 1
= x7 + x6 - x6 + 2x5 - x5 + x4 - x4 + x3 - x3 + x2 - x2 +1
= x2 . ( x5 - x4 + x3 - x + 1 ) + x . ( x5 - x4 + x3 - x + 1 ) + ( x5 - x4 + x3 - x + 1 )
= ( x2 + x + 1 )( x5 - x4 + x3 - x + 1 )
Ta có :
\(x^8+x^7+1\)
\(=\left(x^8+x^7+x^6\right)-x^6+1\)
\(=x^6\left(x^2+x+1\right)-\left[\left(x^3\right)^2-1^2\right]\)
\(=x^6\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6\left(x^2+x+1\right)-\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x^6-\left(x^3+1\right)\left(x-1\right)\right]\)
\(=\left(x^2+x+1\right)\left[x^6-\left(x^4-x^3+x-1\right)\right]\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
Ta có :
x7 + x2 + 1 = x7 - x + x2 + x + 1
= x . ( x3 + 1 ) . ( x3 - 1 ) + ( x2 + x + 1 )
= x . ( x3 + 1 ) . ( x - 1 ) . ( x2 + x + 1 ) + ( x2 + x + 1 )
= ( x2 + x + 1 ) ( x5 - x4 + x2 - x + 1 )
x7 + x2 + 1 = (x7 – x) + (x2 + x + 1)
= x.(x6 – 1) + (x2 + x +1)
= x.(x3 - 1).(x3 +1) + (x2 + x +1)
= x.(x-1).(x2 + x +1).(x3 +1) + (x2+ x +1)
= (x2 + x +1).[x.(x-1).(x3 +1) + 1]
= (x2 + x +1).[(x2-x).(x3 +1) + 1]
= (x2 + x +1).(x5-x4 + x2 -x + 1)