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

Trả lời:

1, 5 ( x + y )² + 15 ( x + y ) 

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

2, ( a - b )² - ( a + b )( b - a )

= ( a - b )² + ( a + b )( a - b )

= ( a - b )( a - b + a + b )

= 2a ( a - b )

Ta có: \(\sqrt{xa}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}\)

\(=\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{a}+\sqrt{b}\right)\)

biểu thức \(\Leftrightarrow\sqrt{a}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{b}\left(\sqrt{x}-\sqrt{y}\right)=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{x}-\sqrt{y}\right)\)

12 - √x - x

= 16 - x - 4 - √x (tách 12 = 16 - 4 và đổi vị trí)

= [42 - (√x)2] - (4 + √x)

= (4 - √x)(4 + √x) - (4 + √x)

= (4 + √x)(4 - √x - 1)

= (4 + √x)(3 - √x)

(với a + b, a - b đều không âm)

xy - y√x + √x - 1

= (√x)2.y - y√x + √x - 1

= y√x(√x - 1) + √x - 1

= (√x - 1)(y√x + 1) với x ≥ 1

1.A

2.C

3.B

4.C

a

c

b

c

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