6: =x^2-7xy+5xy-35y^2

=x(x-7y)+5y(x-7y)

=(x-7y)(x+5y)

7: =x^2-2xy-8xy+16y^2

=x(x-2y)-8y(x-2y)

=(x-2y)(x-8y)

8: =3x^2-6xy-4xy+8y^2

=3x(x-2y)-4y(x-2y)

=(x-2y)(3x-4y)

9: =4x^2+4xy+y^2-16y^2

=(2x+y)^2-16y^2

=(2x+y-4y)(2x+y+4y)

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

10: =2(x^2+5xy+4y^2)

=2(x+y)(x+4y)

11: =5x(x+2y+y^2)

a/ = xy(5y - 6x)

b/ = - (15y² - 2x + 5y - 10xy)

a, x² + x - 6

= x² + 3x - 2x - 6

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

= (x + 3)(x - 2)

b, x² - 2x - 15

= x² - 5x + 3x -15

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

= (x + 3)(x - 5)

c,4x² - 12x - 160

= 4x² - 32x + 20x - 160

= 4x(x - 8) + 20(x - 8)

= (4x + 20)(x - 8)

d, 5x²y - 10xy - 15y

= y(5x² - 10x - 15)

= y(5x² - 15x + 5x - 15)

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

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

Ta có: \(20x=15y=12z\)

\(\Leftrightarrow\dfrac{x}{\dfrac{1}{20}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{12}}\)

\(\Leftrightarrow\dfrac{2x}{\dfrac{1}{10}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{12}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x}{\dfrac{1}{10}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{12}}=\dfrac{y-2x}{\dfrac{1}{15}-\dfrac{1}{10}}=\dfrac{-6}{-\dfrac{1}{30}}=180\)

Do đó: x=9; y=12; z=15

Lời giải:

a) Biểu thức không phân tích được thành nhân tử

b)

\(4x^2-12+8=4x^2-4=4(x^2-1^2)=4(x-1)(x+1)\)

c)

\(9x^2-6xy-3y^2=3(3x^2-2xy-y^2)=3[(3x^2-3xy)+(xy-y^2)]\)

\(=3[3x(x-y)+y(x-y)]=3(x-y)(3x+y)\)

d)

\(25x^2-10xy-15y^2=(5x)^2-2.5x.y+y^2-16y^2\)

\(=(5x-y)^2-(4y)^2=(5x-y-4y)(5x-y+4y)\)

\(=(5x-5y)(5x+3y)=5(x-y)(5x+3y)\)