\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

xin lỗi e mới lớp 7 ak

moi hok lop 6

\(1,a^2-b^2-12a+12b=\left(a-b\right)\left(a+b\right)-12\left(a-b\right)=\left(a-b\right)\left(a+b-12\right)\\ 2,4x^2-4x+1-25y^2=\left(2x-1\right)^2-\left(5y\right)^2=\left(2x-5y-1\right)\left(2x+5y-1\right)\\ c,x^2-3x-10=\left(x^2-5x\right)+\left(2x-10\right)=x\left(x-5\right)+2\left(x-5\right)=\left(x-5\right)\left(x+2\right)\)

1 x mũ 2 + 4xy + 4y mũ 2 = x^2 + 4xy + 4y^2 =(2y+x)^2

2,       4x mũ 2 - 36y mũ 2 =4x^2 -36y^2 = -4 (3y-x) (3y+x)

1, \(x^2+4xy+4y^2=\left(x+2y\right)^2\)

2, \(4x^2-36y^2=\left(2x-6y\right)\left(2x+6y\right)\)

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

4, \(a^3+a^2+3a+3=a^2\left(a+1\right)+3\left(a+1\right)=\left(a^2+3\right)\left(a+1\right)\)

5, \(7x^3-21x^2+3-x=7x^2\left(x-3\right)-\left(x-3\right)=\left(7x^2-1\right)\left(x-3\right)\)

Tưởng tượng nhé :v
(a+b)^2 + 3a+b)+10= t^2 +3t+10 ( đặ a+b=1)  = (t^2+3t+9/4) +31/4 >0  
=> Không thể phân tích :3

1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)

\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)

\(=25\left(a-b\right)^2=25\cdot100=2500\)

2, a^3-3ab^2 = 5

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

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

b^3-3a^2b=10

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

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

=> 100+25 = a^6-6a^4b^2+9a^2b^4+b^6+6a^2b^4+9a^4b^2

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

<=> a^2+b^2 = 5

Khi đó : S = 2016.(a^2+b^2) = 2016.5 = 10080

Tk mk nha

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 : \(a^3-3ab^2-5\Rightarrow\left(a^3-3ab^2\right)^2=25\Rightarrow a^6-6a^4b^2+9a^2b^4=25\)

và \(b^3-3a^2b=10\Rightarrow\left(b^3-3a^2b\right)^2=100\Rightarrow b^6-6b^4a^2+9a^4b^2=100\)

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

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

Nên \(S=2016\left(a^2+b^2\right)=2016.5=10080\)

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

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

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

2,

\(a^3-3ab^2=5\Leftrightarrow\left(a^3-3ab^2\right)^2=25\Leftrightarrow a^6-6a^4b^2+9a^2b^4=25\)

\(b^3-3a^2b=10\Leftrightarrow\left(b^3-3a^2b\right)^2=100\Leftrightarrow b^6-6a^2b^4+9a^4b^2=100\)

\(\Rightarrow a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=125\)

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

\(\Leftrightarrow\left(a^2+b^2\right)^3=125\)

\(\Rightarrow a^2+b^2=5\Rightarrow2016a^2+2016b^2=2016.5=10080\)