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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
\(x^2+x-20\)
\(=x^2+5x-4x-20\)
\(=x\left(x+5\right)-4\left(x+5\right)\)
\(=\left(x-4\right)\left(x+5\right)\)
\(x^2+x-20=\left(x^2+5x\right)-\left(4x+20\right)=x\left(x+5\right)-4\left(x+5\right)=\left(x+5\right)\left(x-4\right)\)
x2 + x - 20
= x2 + 5x - 4x - 20
= ( x2 + 5x ) - ( 4x + 20 )
= x( x + 5 ) - 4( x + 5 )
= ( x - 4 )( x + 5 )
x20 + x +1
= x20 - x2 + x2+ x+ 1
= x2(x18 - 1) + x2+ x+ 1
= x2(x9 + 1)(x9 - 1) + x2+ x+ 1
= x2(x9 + 1)(x3 + 1)(x3 - 1) + x2+ x+ 1
= x2(x9 + 1)(x3 + 1)(x - 1)(x2+ x+ 1) + x2+ x+ 1
= [x2(x9 + 1)(x3 + 1)(x - 1) + 1](x2+ x+ 1)
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Tk cho mình nha
\(\left(x^2-3x+2\right)\left(x^2-9x+20\right)-40=\left(x-1\right)\left(x-2\right)\left(x-4\right)\left(x-5\right)-40\)
\(=\left(x^2-6x+5\right)\left(x^2-6x+8\right)-40\)
Đặt \(t=x^2-6x+5\) thì ta có \(t\left(t+3\right)-40=t^2+3t-40=\left(t+8\right)\left(t-5\right)\)
Suy ra \(\left(x^2-6x+5\right)\left(x^2-6x+8\right)-40=\left(x^2-6x+13\right)\left(x^2-6x\right)=x\left(x-6\right)\left(x^2-6x+13\right)\)
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
x20 + x + 1 = (x20 - x2) + (x2 + x + 1)
= x2(x18 - 1) + (x2 + x + 1)
= x2(x9 - 1)(x9 + 1) + (x2 + x + 1)
=(x11 + x)(x3 - 1)(x6 + x3 + 1) + (x2 + x + 1)
= (x17 + x14 + x11 + x7 + x4 + x)(x - 1)(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x18 + x15 + x12 + x8 + x5 + x2 - x17 - x14 - x11 - x7 - x4 - x + 1)