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

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

Ai trả lời mình đi cho đỡ quê 😞💔

a: \(=\left(x+2-y\right)\left(x+2+y\right)\)

c: \(=\left(x-y\right)^2\)

d: \(=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

a) Kết quả 2x(2x – 3).             b) Kết quả xy( x 2  – 2xy + 5).

c) Kết quả 2x(x + 1)(x + 4).    d) Kết quả 2 5 ( y − 1 ) ( x + y ) .

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+21\)

\(=\left(x^2+5x+3\right)\left(x^2+5x+7\right)\)

a, \(\left(x+y\right)^3+\left(x-y\right)^3=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

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

b, \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)

Đặt \(x^2+5x+5=t\)

\(\left(t-1\right)\left(t+1\right)-3=t^2-4=\left(t-2\right)\left(t+2\right)\)

Theo cách đặt \(\left(x^2+5x+3\right)\left(x^2+5x+7\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

a: =(x-y)(5-y)

b: \(=x^2-6x+9-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

\(a,5\left(x-y\right)-y\left(x-y\right)=\left(5-y\right)\left(x-y\right)\\ b,x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)