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

\(x^2+x-6=x^2+2.\frac{1}{2}x+\frac{1}{4}-\frac{25}{4}=\left(x+\frac{1}{2}\right)^2-\frac{25}{4}=\left(x+\frac{1}{2}-\frac{5}{2}\right)\left(x+\frac{1}{2}+\frac{5}{2}\right)=\left(x-2\right)\left(x+3\right)\)

x²+x-6=x³+3x-2x-6=x(x+3)-2(x+3)=(x+3)(x-2)

x²+x-6=\(x^2+x+\frac{1}{4}-6-\frac{1}{4}=\left(x+\frac{1}{2}\right)^2-\left(\frac{5}{2}\right)^2=\left(x+\frac{1}{2}-\frac{5}{2}\right)\left(x+\frac{1}{2}+\frac{5}{2}\right)=\left(x-2\right)\cdot\left(x+3\right)\)

\(1,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)

\(3,7x-6x^2-2=-6x^2+7x-2=-6x^2+3x+4x-2=3x\left(-2x+1\right)+2\left(2x-1\right)\)

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

\(2,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)

cách 1 :  \(x^2\)\(-\)\(5x\)\(+\)\(6\)

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

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

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

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

Cách 2 :  \(x^2\)\(-\)\(5x\)\(+\)\(6\)

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

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

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

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

Mình chỉ biết một cách thôi

x^2-5x+6

=x^2-2x-3x+6

=x(x-2)-3(x-2)

=(x-3)(x-2)

\(2x^2-5x+2\)

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

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

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

 =x³-7x+6

=x³-2x²+2x²-4x-3x+6

=x²(x-2)+2x(x-2)-3(x-2)

=(x-2)(x²+2x-3)

=(x-2)(x²+2x+1-4)

=(x-2)[(x+1)²-4]

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

x³ - 7x + 6

= x³ - 2x² + 2x² - 4x - 3x + 6

= x² ( x - 2 ) + 2x ( x - 2 ) + 3 ( x - 2 )

= ( x² + 2x + 3 ) ( x - 2 )

= ( x² + 2x + 1 - 4 ) ( x - 2 )

= [ ( x + 1 )² - 2² ] ( x - 2 )

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

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

a)   \(\left(x-7\right)\left(x+2\right)\)

b)  \(\left(2x-3\right)\left(x+2\right)\)

a) Ta có : x² + 7x + 12 

= x² + 3x + 4x + 12

= (x² + 3x) + (4x + 12)

= x(x + 3) + 4(x + 3)

= (x + 4)(x + 3)

Bạn ơi mk nhầm đề rồi số 30 thay bằng số 60 còn 36 thay bằng 72 và 39 thay bằng 75 nha 

Ta có \(x^2+7x+12\)

=\(x^2+3x+4x+12\)

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

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

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