a. \(=4x^3-12x^2-x^2+3x+6x-18=\left(x-3\right)\left(4x^2-x+6\right)\)

b.  \(=-x^3+x^2-7x^2+7x-x+1=\left(x-1\right)\left(-x^2-7x-1\right)\)

c.  \(=x^3+2x^2-6x^2-12x+4x+8=\left(x+2\right)\left(x^2-6x+4\right)\)

a, = (x^3-x^2)-(4x^2-4x)+(4x-4)

    = (x-1).(x^2-4x+4) = (x-1).(x-2)^2

b, = (x^3+x^2)-(10x^2+10x)+(16x+16)

 = (x+1).(x^2-10x+16)

 = (x+1).[ (x^2-2x)-(8x-16) ] = (x+1).(x-2).(x-8)

k mk nha

a)= (x^3-x^2)-(4x^2-4x)+(4x-4)

    = (x-1).(x^2-4x+4)

    = (x-1).(x-2)^2

b)= (x^3+x^2)-(10x^2+10x)+(16x+16)

 = (x+1).(x^2-10x+16)

 = (x+1).[ (x^2-2x)-(8x-16) ]

 = (x+1).(x-2).(x-8)

P/s tham khảo nha

a) \(4x^2+4x-3\)

\(=4x^2-2x+6x-3\)

\(=2x\left(2x-1\right)+3\left(2x-1\right)\)

\(=\left(2x+3\right)\left(2x-1\right)\)

b) \(2x^2+xy-y^2\)

\(=2x^2+2xy-xy-y^2\)

\(=2x\left(x+y\right)-y\left(x+y\right)\)

\(=\left(2x-y\right)\left(x+y\right)\)

c) \(x^2+9x-22\)

\(=x^2+11x-2x-22\)

\(=x\left(x+11\right)-2\left(x+11\right)\)

\(=\left(x-2\right)\left(x+11\right)\)

a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)

b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)

c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)

d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)

a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)

\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)

\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)

\(=3\left(x-2\right)\left(x+5\right)\)

c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)

\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)

d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)

\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)

a,\(x^2-5x+6=x^2-2x-3x+6 =x(x-2)-3(x-2) =(x-2)(x-3)\)

b,\(3x^2+9x-30=3x^2+15x-6x-30=3x(x+5)-6(x+5)=(x+5)(3x-6)\)

c,\(x^2-7x+12=x^2-4x-3x+12=x(x-4)-3(x-4)=(x-4)(x-3)\)

d,\(x^2-7x+10=x^2-2x-5x+10=x(x-2)-5(x-2)=(x-2)(x-5)\)

Hok tốt !!!!!!!!!!

\(a,x^2-5x+6\\=x^2-3x-2x+6\\=x(x-3)-2(x-3)\\=(x-3)(x-2)\\---\\b,3x^2+9x-30\\=3x^2-6x+15x-30\\=3x(x-2)+15(x-2)\\=(x-2)(3x+15)\\=3(x-2)(x+5)\\---\)

\(c,x^2-3x+2\\=x^2-x-2x+2\\=x(x-1)-2(x-1)\\=(x-1)(x-2)\\---\\d,3x^2-5x-2\\=3x^2-6x+x-2\\=3x(x-2)+(x-2)\\=(x-2)(3x+1)\\Toru\)

a)   \(x^2-3x+2\)

\(\Leftrightarrow x^2-2x-x+2\)

\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\)

b)  \(x^2-6x+8\)

\(\Leftrightarrow x^2-4x-2x+8\)

\(\Leftrightarrow x\left(x-4\right)-2\left(x-4\right)\)

\(\Leftrightarrow\left(x-4\right)\left(x-2\right)\)

c)  \(3x^2+9x-30\)

\(\Leftrightarrow3\left(x^2+3x-10\right)\)

\(\Leftrightarrow3\left[\left(x^2+2\cdot\frac{3x}{2}+\frac{9}{4}\right)-\frac{49}{4}\right]\)

\(\Leftrightarrow3\left[\left(x+\frac{3}{2}\right)^2-\left(\frac{7}{2}\right)^2\right]\)

\(\Leftrightarrow3\left(x+\frac{3}{2}+\frac{7}{2}\right)\left(x+\frac{3}{2}-\frac{7}{2}\right)\)

\(\Leftrightarrow3\left(x-2\right)\left(x+5\right)\)

d)  \(x^2-9x+18\)

\(\Leftrightarrow x^2-3x-6x+18\)

\(\Leftrightarrow x\left(x-3\right)-6\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)

TK MK NKA !!!! TH@NK !!! 

a) x² - 3x + 2 ( như này mới phân tích được ạ :) )

= x² - x - 2x + 2

= x( x - 1 ) - 2( x - 1 )

= ( x - 2 )( x - 1 )

b) x² - 6x + 8

= x² - 2x - 4x + 8

= x( x - 2 ) - 4( x - 2 )

= ( x - 4 )( x - 2 )

c) 3x² + 9x - 30

= 3( x² + 3x - 10 )

= 3( x² - 2x + 5x - 10 )

= 3[ x( x - 2 ) + 5( x - 2 )]

= 3( x + 5 )( x - 2 )

d) x² - 9x + 18

= x² - 3x - 6x + 18

= x( x - 3 ) - 6( x - 3 )

= ( x - 6 )( x - 3 )

a, \(x^2-3x+2=x^2-x-2x+2\)

\(=x\left(x-1\right)-2\left(x-1\right)=\left(x-2\right)\left(x-1\right)\)

b, \(x^2-6x+8=x^2-2x-4x+8\)

\(=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

c, \(3x^2+9x-30=3\left(x^2+3x-10\right)\)

\(=3\left(x^2-2x+5x-10\right)=3\left(x+5\right)\left(x-2\right)\)