1) \(x^2+6xy+5y^2-5y-x\)

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

\(=x\left(x+y-1\right)+5y\left(x+y-1\right)\)

\(\left(x+5y\right)\left(x+y-1\right)\)

2) Ta có : \(a^3-3ab^2=5\)

\(\Rightarrow\)\(\left(a^3-3ab^2\right)^2-100=25\Rightarrow a^6-6a^4b^2+9a^2b^4=25\)

Và \(b^3-3a^2b=10\)

\(\Rightarrow\)\(\left(b^3-3a^2b\right)^2=100\Rightarrow b^6-6b^4a^2-9a^4b^2=100\)

\(\Rightarrow\)\(125=a^6+b^6+3a^2b^4+3a^4b^2\)

Hoặc \(125=\left(a^2+b^2\right)^3\Rightarrow a^2+b^2=5\)

Do đó : \(S=2016\left(a^2+b^2\right)=2016.5=10080\)

= x² + 6xy + 5y² - 5y - x

= 5y² + 5xy - 5y + xy + x² - x

= 5y(y + x - 1) + x(y + x - 1)

= (y + x - 1)(5y + x)

Bài 2 :

1) \(x^2+6xy+5y^2-5y-x=x^2-x+xy+5y^2-5y+5xy\)

\(=x\left(x-1+y\right)+5y\left(y-1+x\right)=\left(x+y-1\right)\left(x+5y\right)\)

Ca ca câu này mụi lm đc òi, lm hộ mụi mấy cái khác ik

đề bạn ra hình như sai thì phải 

1, x^2 + 6xy + 5y^2 - 5y - x 

= x^2 + xy - x + 5xy + 5y^2 - 5y

= x(x + y - 1) + 5y(x + y - 1)

= (x + 5y)(x + y - 1

2, 

a^3 - 3ab^2 = 5

<=> (a^3 - 3ab^2)^2 = 25

<=> a^6 - 6a^4b^2 + 9a^2b^4 = 25      (1)

b^3 - 3a^2b = 10

<=> (b^3 - 3a^2b)^2 = 100

<=> b^6 - 6b^4a^2 + 9a^4b^2 = 100     (2)

(1) + (2) = a^6 - 6a^4b^2 + 9a^2b^4  + b^6 - 6b^4a^2 + 9a^4b^2 = 25 + 100

<=> a^6 + 3a^4b^2 + 3a^2b^4 + b^6 = 125

<=> (a^2 + b^2)^3 = 125

<=> a^2 + b^2 = 5 

<=> 2016(a^2 + b^2) = 5.2016

<=> 2016a^2 + 2016b^2 = 10080

a. x² + 7xy + 10y²

= x² + 2xy + 5xy + 10y²

= x.(x + 2y) + 5y.(x + 2y)

= (x + 2y).(x + 5y)

b. x² - 6xy + 5y²

= x² - xy - 5xy + 5y²

= x.(x - y) - 5y.(x - y)

= (x - y).(x - 5y)