a.

\(5x^2\left(x-2y\right)-15x\left(x-2y\right)\)

\(=\left(x-2y\right)\left(5x^2-15x\right)\)

\(=5x\left(x-2y\right)\left(x-3\right)\)

b. 

\(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(3+5x\right)\)

\(a,5x^2\left(x-2y\right)-15x\left(x-2y\right)\) 

\(=5x\left(x-2y\right)\left(x-3\right)\) 

\(b,3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\) 

\(=\left(x-y\right)\left(3+5x\right)\)

Chúc bạn học tốt!

a) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=\left(x-2y\right)\left(5x^2-15x\right)=\left(x-2y\right)5x\left(x-3\right).\)
\(b.3\left(x-y\right)-5\left(y-x\right)=3\left(x-y\right)+5\left(x-y\right)\)
    \(=\left(3+5\right)\left(x-y\right)=8\left(x-y\right)\)

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)

=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

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

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

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

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

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

5: \(\left(3x-y\right)^2-4y^2\)

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

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

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

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

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

\(=\left(2x-1\right)^2-\left(3y\right)^2\)

\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

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

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

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

a) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=\left(5x^2-15x\right)\left(x-2y\right)=5x\left(x-3\right)\left(x-2y\right)\)

b) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(3+5x\right)\left(x-y\right)\)

14.C

15.D

Câu 14: C

Câu 15: D

14:A
15:D

A. Cách B sai vì 5 : 2/5 thì ko thể nào = 25 đc.

Answer:

Câu đầu bạn xem lại.

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

\(=\left(3x\right)^2+2.2x.4+4^2+\left(4x\right)^2-2.4x.3+3^2+2^2-\left(5x\right)^2\)

\(=9x^2+24x+16+16x^2-24x+9+4-25x^2\)

\(=\left(9x^2+16x^2-25x^2\right)+\left(24x-24x\right)+\left(16+9+4\right)\)

\(=29\)

\(\left(5x+y\right).\left(25x^2-5xy+y^2\right)-\left(5x-y\right).\left(25x^2+5xy+y^2\right)\)

\(=\left(5x+y\right).[\left(5x\right)^2-5x.y+y^2]-\left(5x-y\right).[\left(5x\right)^2+5x.y+y^2]\)

\(=\left(5x\right)^3+y^3-[\left(5x\right)^3-y^3]\)

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

\(=2y^3\)

a) 3( x - y ) - 5x( y - x )

= 3( x - y ) - 5x[ -( x - y ) ]

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

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

b) x³ + 2x²y + xy² - 9x

= x( x² + 2xy + y² - 9 )

= x[ ( x + y )² - 3² ]

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

c) 14x²y - 21xy² + 28x²y²

= 7xy( 2x - 3y + 4xy )

                                              Bài giải

\(a,\text{ }3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(3+5x\right)\)

\(b,\text{ }x^3+2x^2y+xy^2-9x\)

\(=x\left(x^2+2xy+y^2-9\right)\)

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

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

\(c,\text{ }14x^2y-21xy^2+28x^2y\)

\(=7xy\left(2x-3y+4x\right)\)

\(=7xy\left(6x-3y\right)\)

a, \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(3+5x\right)\)

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

c, \(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)