Ta có x 2 – 7x + 10

= x 2 – 2x – 5x + 10

= x(x – 2) – 5(x – 2) = (x – 5)(x – 2)

Đáp án cần chọn là: B

\(a,=\left(x-y\right)\left(x+y\right)+11\left(x-y\right)=\left(x-y\right)\left(x+y+11\right)\\ b,=\left(x+z\right)\left(x^2-xz+z^2\right)+y\left(x^2+z^2-xz\right)\\ =\left(x^2-xz+z^2\right)\left(x+y+z\right)\)

a. x² - y² + 11x - 11y

= (x + y)(x - y) + 11(x - y)

= (x + y + 11)(x - y)

b. Mik ko hiểu đề lắm

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

a) \(\left(x^2-7x+12\right).\left(x^2-15x+56\right)-60\)

\(=\left(x-3\right)\left(x-4\right)\left(x-7\right)\left(x-8\right)-60\)

b) \(x^4+2000x^2+1999x+2000\)

\(=\left(x^2-x+2000\right)\left(x^2+x+1\right)\)

\(=\left(x^2+x-1\right)^2+1999\left(x^2+x+1\right)+1\)

hình như Thành lộn đề rồi đấy ạ, ý mình là phân tích thành nhân tử á

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

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

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

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

a) x² - 5x + 6 = (x²-2x)-(3x-6)=x(x-2)-3(x-2)=(x-3)(x-2)

b) 3x² + 9x -30= 3(x²+3x-10) = 3((x²+5x)-(2x+10)) = 3(x(x+5)-2(x+5)) = 3(x-2)(x+5)

c) x² + 7x + 10 =( x²+5x)+(2x+10)=x(x+5)+2(x+5)=(x+2)(x+5)

a) (x - 2)(x - 3).                        b) 3(x - 2)(x + 5).

c) (x - 2)(3x + 1).                     d) (x-2y)(x - 5y).

e) (x + l)(x + 2)(x - 3).             g) (x-1)(x + 3)( x 2  + 3).

h) (x + y - 3)(x - y + 1).

a. \(x^2\) - 9y²

= (\(x\))² - (3y)²

= (\(x\) - 3y)(\(x\) + 3y)

x² - 9y² = x² - (3y)²= (x - 3y)(x + 3y)

a: =3x^2-3x-8x+8=(x-1)(3x-8)

b: =x^2-x-5x+5=(x-1)(x-5)

c: =x^2-6x+2x-12=(x-6)(x+2)