Ta có:x(3x²+4x-7)=x[(3x²-3x)+(7x-7)]=x[3x(x-1)+7(x-1)]=x(x-1)(3x+7)

\(3x^4+6x^3-7x^2+8x-10\)

\(=\left(3x^4-3x^3\right)+\left(9x^3-9x^2\right)+\left(2x^2-2x\right)+\left(10x-10\right)\)

\(=\left(x-1\right)\left(3x^3+9x^2+2x+10\right)\)

Bài này đúng đề không

Dễ mà ^_^:  3x²-7x+2=3x² -x-6x+2=(3x²-x)-(6x-2)=x(3x-1)-2(3x-1)=(3x-1)(x-2) 

\(3x^2-7x+2=\left(3x^2-6x\right)-\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)=\left(3x-1\right)\left(x-2\right)\)

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

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

b: \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

\(=5x\left(x+y\right)+y\left(x+y\right)\)

\(=\left(x+y\right)\left(5x+y\right)\)

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

d: \(4x^2+12x+9-y^2\)

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

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

\(3x^4+11x^3-7x^2-2x+1=\left(3x^4+12x^3-3x^2-3x\right)+\left(-x^3-4x^2+x+1\right)\)

\(=\left(3x-1\right)\left(x^3+4x^2-x-1\right)\)

\(=2x^4+6x^3-3x^3-9x^2-3x^2-9x+2x+6\)

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

\(=\left(x+3\right)\left(2x^3-4x^2+x^2-2x-x+2\right)=\left(x+3\right)\left(x-2\right)\left(2x^2+x-1\right)\)

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

2x^4+3x^3-12x^2-7x+6 = (2x^4-x^3)+(4x^3-2x^2)-(10x^2-5x)-(12x-6)

= x^3.(2x-1)+2x^2.(2x-1)-5x.(2x-1)-6.(2x-1) = (2x-1).(x^3+2x^2-5x-6) 

= (2x-1).[ (x^3+x^2)+(x^2+x)-(6x+6) ] = (2x-1).(x+1).(x^2+x-6) = (2x-1).(x-1).[(x^2-2x)+(3x-6)]

= (2x-1).(x+1).(x-2).(x+3)

k mk nha

\(2x^4+3x^3-12x^2-7x+6\)

\(=\left(2x^4+2x^3\right)+\left(x^3+x^2\right)-\left(13x^2+13x\right)+\left(6x+6\right)\)

\(=\left(x+1\right)\left(2x^3+x^2-13x+6\right)\)