bài 1
\(\text{a) 6x^2 – 3xy = 3x( 2x – y )}\)
\(\text{b) x^2 – y^2 – 6x + 9 = x^2 – 6x + 9 – y^2 = ( x – 3 )^2 = ( x – 3 – y )( x – 3 + y )}\)
\(\text{c) x^2 + 5x – 6 = x^2 – x + 6x – 6 = ( x – 1 )( x + 6 )}\)
bài 2
\(\text{a) ( x + 2 )^2 – ( x – 3 )(x + 1 ) = x^2 + 4x + 4 –x^2 +2x +3 = 6x + 7}\)
\(\text{b) x^3 -2x^2 + 5x – 10 = x^2 ( x – 2 ) +5 ( x – 2 ) = ( x – 2 )( x^2 + 5 )}\)
2:
a: =x^2+4x+4-(x^2+x-3x-3)
=x^2+4x+4-(x^2-2x-3)
=x^2+4x+4-x^2+2x+3
=6x+7
b: \(\dfrac{x^3-2x^2+5x-10}{x-2}=\dfrac{x^2\left(x-2\right)+5\left(x-2\right)}{x-2}\)
=x^2+5
1:
a: 6x^2-3xy
=3x*2x-3x*y
=3x(2x-y)
b: x^2-y^2-6x+9
=(x^2-6x+9)-y^2
=(x-3)^2-y^2
=(x-3-y)(x-3+y)
c: x^2+5x-6
=x^2+6x-x-6
=(x+6)(x-1)
\(B1:\\ a,6x^2-3xy=3x\left(2x-y\right)\\---\\ b,x^2-y^2-6x+9\\ =\left(x^2-6x+9\right)-y^2\\ =\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\\ c,x^2+5x-6\\ =x^2+6x-x-6\\ =x\left(x+6\right)-x\left(x+6\right)\\ =\left(x-1\right)\left(x+6\right)\)
Bài 2:
\(a,\left(x+2\right)^2-\left(x-3\right)\left(x+1\right)\\ =\left(x^2+4x+4\right)-\left(x^2-3x+x-3\right)\\ =x^2-x^2+4x+3x-x+4+3\\ =6x+7\\ b,\left(x^3-2x^2+5x-10\right):\left(x-2\right)\\ =\left(x^3-2x^2+5x-10\right):x-\left(x^3-2x^2+5x-10\right):2\\ =x^2-2x+5-\dfrac{10}{x}-\dfrac{1}{2}x^3-x^2+\dfrac{5}{2}x-5\\ =-\dfrac{1}{2}x^3+\dfrac{1}{2}x-\dfrac{10}{x}\)
