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

Ta có:

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

\(=\left(x+y\right)^5-\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)

\(=\left(x+y\right)\left[\left(x+y\right)^4-x^4+x^3y-x^2y^2+xy^3-y^4\right]\)

\(=\left(x+y\right)\left[x^4+4x^3y+6x^2y^2+4xy^3+y^4-x^4+x^3y-x^2y^2+xy^3-y^4\right]\)

\(=\left(x+y\right)\left(5x^3y+5x^2y^2+5xy^3\right)=5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)

A = ( x + y )⁵ - x⁵ - y⁵

A = ( x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵ ) - x⁵ - y⁵

A = 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴

A = 5xy( x³ + 2x²y + 2xy² + y³ )

A = 5xy[ ( x + y )( x² - xy + y² ) + 2xy( x + y ) ]

A = 5xy( x + y )( x² + xy + y² )

\(\left(x+y\right)^5-x^5-y^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5-x^5-y^5=5x^4y+10x^3y^2+10x^2y^3+5xy^4=5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)

để lâu cứt trâu hoá bùn

Thằng ngáo lol

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

sao chép à bạn :)))