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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
\(x^2-6x+8\)
\(=\left(x^2-6x+9\right)-1\)
\(=\left(x-3\right)^2-1^2\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
(Tíck cho mìk vs nha!)
cách 2:
x2 -6x +8 = x2 -2x -4x+8= x(x-2) -4(x-2)
= (x-2)(x-4)
\(x^2+x-2\)
\(=x^2-x+2x-2\)
\(=x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
chì làm được 1 cách thôi
Có : x^2-x-6 = (x^2-3x)+(2x-6) = x.(x-3)+2.(x-3) = (x-3).(x+2)
Tk mk nha
\(2x^2+x-6\)
\(=2x^2+4x-3x-6\)
\(=2x\left(x+2\right)-3\left(x+2\right)\)
\(=\left(2x-3\right)\left(x+2\right)\)
\(x.\left(x^2-4\right)-3x+6\)
\(=x.\left(x+2\right).\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x^2+2x\right).\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2+2x-3\right)\)
\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)
\(=\left(x-2\right).[x.\left(x-1\right)+3.\left(x-1\right)]\)
\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)
Ta có : x^2 + x - 6
= x^2 + 3x - 2x - 6
= x ( x + 3 ) - ( 2x + 6 )
= x ( x + 3 ) - 2 ( x + 3 )
= ( x - 2 ) ( x + 3 )
sửa lại
x^2+x-6
=x2+2x-3x-6
=x(x+2)-3(x+2)
=(x+2)(x-3)
\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3.\left(x-2\right)\)
\(=\left(x+3\right)\left(x-2\right)\)