a) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)

\(=2x4x^2+2x2x+2x-4x^2-2x-1\)

\(=8x^3+4x^2+2x-4x^2-2x-1\)

\(=8x^3-1\)

b) \(\left(x+2y+z\right)\left(x+2y-z\right)\)

\(=x^2+2xy-xz+2xy+4y^2-2yz+xz+2yz-z^2\)

\(=x^2+2xy+2xy+4y^2-z^2\)

c)\(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)

\(=x^6+3x^4+9x^2-3x^4-9x^2-27\)

\(=x^6-27\)

Áp dụng hằng đẳng thức \(\left(a+b\right)\left(a-b\right)=a^2-b^2\)ta đc:

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

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

                                                             \(=x^2+4xy+4y^2-z^{ }\)

( x + 2y + z ) ( x + 2y - z ) 

<=> ( x + 2y)² - z² 

<=> x² + 4xy + 4y² - z² 

^^ Học tốt! 

Cái cuối là z² nha bn

Giúp nhanh giùm mình với ạ 😭😭😭😭😭😭

a) ( 3x + 2y - 1 )( x - 5 ) - ( x - 2 )2y 

= 3x(x - 5) + 2y(x - 5) - 1(x - 5) - ( 2xy - 4y )

= 3x² - 15x + 2xy - 10y - x + 5 - 2xy + 4y

= 3x² - 16x - 6y + 5 

b) ( 3x - 2 )( 3x + 2 ) - ( 2x + 1 )( 4x + 3 ) 

= [ ( 3x )² - 2² ] - ( 8x² + 10x + 3 )

= 9x² - 4 - 8x² - 10x - 3

= x² - 10 - 7

a: \(=\dfrac{3x+6-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+10}{x^2-4}\)

b: \(=\dfrac{10x+15+4x-6+2x+5}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x+14}{\left(2x-3\right)\left(2x+3\right)}\)

Answer:

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

\(=3x^2.2x^3+3x^2.(-x)+3x^2.5\)

\(=6x^5-3x^3+15x^2\)

\((4xy+3y-5x).x^2y\)

\(=4xy.x^2y+3y.x^2y-5x.x^2y\)

\(=4x^3+3x^2y^2-5x^3y\)

Bài 1: 

a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)

= \(x^2\) -  16 - 5\(x\) - 5 + \(x^2\) + \(x\) 

= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)

= 2\(x^2\) - 4\(x\) - 21

b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)

=  3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7

= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)

= - 3\(x^2\) + 3\(xy\) - 3

Bài 2:

a, 3\(x^2\).(2\(x\) + y) - 2y(4\(x^2\) - y)

= 6\(x^3\) + 3\(x^2\).y - 8y\(x^2\) + 2y²

= 6\(x^3\) - (8\(x^2\)y - 3\(x^2\)y) + 2y²

= 6\(x^3\) - 5\(x^2\)y + 2y²

a/ (x-1)²-(4x+3)(2-x)=x²-2x+1-(8x-4x²+6-3x)

=x²-2x+1-8x+4x²-6+3x=5x²-7x-6

b/ (15x³y² - 6x²y³) : 3x²y² = 5x - 2y

c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)

a) Ta có: ( x2 -1 )( x2 + 2x )

= x2( x2 + 2x ) - ( x2 + 2x )

= x4 + 2x3 - x2 - 2x

b) Ta có ( x + 3 )( x2 + 3x -5 )

= x( x2 + 3x -5 ) + 3( x2 + 3x -5 )

= x3 + 3x2 - 5x + 3x2 + 9x - 15

= x3 + 6x2 + 4x - 15

c) Ta có ( x -2y )( x2y2 - xy + 2y )

= x( x2y2 - xy + 2y ) - 2y( x2y2 - xy + 2y )

= x3y2 - x2y + 2xy - 2x2y3 + 2xy2 - 4y2

d) Ta có ( 1/2xy -1 )( x3 -2x -6 )

= 1/2xy( x3 -2x -6 ) - ( x3 -2x -6 )

= 1/2x4y - x2y - 3xy - x3 + 2x + 6