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

a, \(2xy-x^2+49-y^2=49-\left(x^2-2xy+y^2\right)\)

                                          \(=7^2-\left(x-y\right)^2\)

                                          \(=\left(7-x+y\right)\left(7+x-y\right)\)

b, \(x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)

Bài này dễ mà. Chúc bạn học tốt.

a) \(2xy-x^2+49-y^2\)

\(=49-\left(x^2-2xy+y^2\right)\)

\(=49-\left(x-y\right)^2\)

\(=\left(7+x-y\right)\left(7-x+y\right)\)

b) \(x^3+2x^2+x\)

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

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

Bài 1:

a: \(5x^3+10xy=5x\left(x^2+2y\right)\)

b: \(x^2+14x+49-y^2\)

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

\(=\left(x+7+y\right)\left(x+7-y\right)\)

trả lời 

e chưa học đến nha 

nếu ko a lên học 24 hỏi nha 

chúc a học tốt

#)Giải :

a, 16( x - y )² - 49( x + y )²

= [4( x - y )]² - [7( x + y )]²

= [ 4x - 4y ]² - [ 7x + 7y ]²

= ( 4x - 4y + 7x + 7y )[( 4x - 4y ) - ( 7x + 7y )]

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

           #~Will~be~Pens~#

a) \(16\left(x-y\right)^2-49\left(x+y\right)^2\)

\(=\left[4\left(x-y\right)\right]^2-\left[7\left(x+y\right)\right]^2\)

\(=\left[4\left(x-y\right)-7\left(x+y\right)\right]\left[4\left(x-y\right)+7\left(x+y\right)\right]\)

\(=\left(4x-4y-7x-7y\right)\left(4x-4y+7x+7y\right)\)

\(=\left(-3x-11y\right)\left(11x+3y\right)\)

\(a,49.\left(y-4\right)^2-9y^2-36y-36=49\left(y-4\right)^2-9\left(y^2+4y+4\right)\)

\(=49\left(y-4\right)^2-9\left(y+4\right)^2=\left(7y-28\right)^2-\left(3y+12\right)^2\)

\(=\left(7y-28+3y+12\right)\left(7y-28-3y-12\right)\)

\(=\left(10y-16\right)\left(4y-40\right)=8\left(5y-8\right)\left(y-10\right)\)

\(b,xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)

\(=xyz-xy-yz-xz+x+y+z-1\)

\(=\left(xyz-xy\right)-\left(xz-x\right)-\left(yz-y\right)+\left(z-1\right)\)

\(=xy\left(z-1\right)-x\left(z-1\right)-y\left(z-1\right)+\left(z-1\right)\)

\(=\left(z-1\right)\left(xy-x-y+1\right)\)

\(=\left(z-1\right)\text{[}x\left(y-1\right)-\left(y-1\right)\text{]}\)

\(=\left(z-1\right)\left(y-1\right)\left(x-1\right)\)

=(7y-3x-34)(7y+3x-22)

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)