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

1D  2C

Câu 1: D

Câu 2: C

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

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

1. 5x² - 4x

= x(5x - 4)

2. 8x²(x - 3y) - 12x(x - 3y)

= (8x² - 12x)(x - 3y)

= 4x(2x - 3)(x - 3y)

3. 3(x - y) - 5x(y - x)

= 3(x - y) + 5x(x - y)

= (3 + 5x)(x - y)

bấm máy tính là ra đó :VV

b, x mũ 2 + 4x + 3 = ( x+ 1 ) ( x + 3 )

a) = x² + 3x + x + 3 = x( x + 3 ) + ( x + 3 ) = ( x + 3 )( x + 1 )

b) = x² - 3x - 5x + 15 = x( x - 3 ) - 5( x - 3 ) = ( x - 3 )( x - 5 )

c) = x² - 2x + 4x - 8 = x( x - 2 ) + 4( x - 2 ) = ( x - 2 )( x + 4 )

d) = x² - 4x + 3x - 12 = x( x - 4 ) + 3( x - 4 ) = ( x - 4 )( x + 3 )

e) = 4x² - 2x + 6x - 3 = 2x( 2x - 1 ) + 3( 2x - 1 ) = ( 2x - 1 )( 2x + 3 )

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

\(=4x\left(x-y\right)+3\left(x-y\right)\left(x-y\right)\)

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

\(=\left(x-y\right)\left(4x+3x-3y\right)\)

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

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

\(=\)\(4x\left(x-y\right)+\left(3x-3y\right)\left(x-y\right)\)

\(=\)\(\left(x-y\right)\left(4x+3x-3y\right)\)

\(=\)\(\left(x-y\right)\left(7x-3y\right)\)

Chúc bạn học tốt ~ 

Ta có:

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

\(=4x\left(x-y\right)+3\left(x-y\right)\left(x-y\right)\)

\(=\left(x-y\right)\left(4x+3x-3y\right)\)

\(=\left(x-y\right)\left(7x-3y\right)\)

1.A

2.C

3.B

4.C

a

c

b

c

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

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

4x( y - x ) + 7y( x - y ) + y²( y - x )

= 4x( y - x ) - 7y( y - x ) + y²( y - x )

= ( y - x )( 4x - 7y + y² )