Bài giải:

a) (-2x⁵ + 3x² – 4x³) : 2x² = (- $\frac{2}{2}$)x^{5 – 2} + $\frac{3}{2}$x^{2 – 2} + (-$\frac{4}{2}$)x^{3 – 2} = - x³ + $\frac{3}{2}$ – 2x.

b) (x³ – 2x²y + 3xy²) : (- $\frac{1}{2}$x) = (x³ : -$\frac{1}{2}$x) + (-2x²y : -$\frac{1}{2}$x) + (3xy² : -$\frac{1}{2}$x)

= -2x² + 4xy – 6y²

c)(3x²y² + 6x²y³ – 12xy) : 3xy = (3x²y² : 3xy) + (6x²y² : 3xy) + (-12xy : 3xy)

= xy + 2xy² – 4.

a) (-2x⁵+3x²-4x³) : 2x²

= (-2x⁵:2x²)-(4x³:2x²)+(3x²:2x²)

= -x³-2x+\(\dfrac{3}{2}\)

b) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)

= \(\left(x^3:\dfrac{-1}{2}x\right)+\left(-2x^2y:\dfrac{-1}{2}x\right)+\left(3xy^2:\dfrac{-1}{2}x\right)\)

= \(-2x^2+4xy-6y^2\)

c) \(\left(3x^2y^2+6x^2y^3-12xy\right):3xy\)

= \(\left(6x^2y^3:3xy\right)+\left(3x^2y^2:3xy\right)+\left(-12xy:3xy\right)\)

= \(xy^2+xy-4\)

a: \(=\left(4xy^2+2xy^2\right)+\left(3x^2y-3x^2y\right)=6xy^2\)

b: \(=xy\left(\dfrac{1}{5}+\dfrac{1}{3}\right)+xy^2\left(\dfrac{4}{3}-\dfrac{2}{5}\right)=\dfrac{8}{15}xy+\dfrac{14}{15}xy^2\)

d: \(=\dfrac{-4}{9}\cdot\dfrac{3}{2}\cdot xy^2\cdot xy^3=-\dfrac{2}{3}x^2y^5\)

a ) ( x - 2 )( x + 5 )

= x^2 + 5x - 2x + 10

= x^2 + 3x + 10

b ) 3x + 3y +ax + ay 

= x( 3 + a ) + y( 3 + a )

= ( 3 + a )( x + y ) 

c ) ( x^2 + 2xy ) : ( x + 2y )

= [ x( x + 2y ) ] : ( x + 2y )

= x  : 1

= x 

d ) ( x - 2 )( x + 2 ) + ( x + 1 )^2 - 2x^2 = 0 

       x^2 + 2x - 2x - 4 + x^2 + x + x + 1 - 2x^2 = 0

       x^2 - 4 + x^2 + 2x + 1 - 2x^2 = 0

        2x^2 + 2x - 4 + 1 - 2x^2 = 0 

       2x - 3 = 0

       2x = 0 + 3

       2x = 3

        x = 3 : 2

        x = 3/2 

a)   \(\left(x-2\right)\left(x+5\right)\)

\(=x^2+5x-2x-10\)

\(=x^2+3x-10\)

b)  \(3x+3y+ax+ay\)

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

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

c)   \(\left(x^2+2xy\right):\left(x+2y\right)\)

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

\(=x\)

d)  \(\left(x-2\right)\left(x+2\right)+\left(x+1\right)^2-2x^2=0\)

\(\Leftrightarrow\)\(x^2-4+x^2+2x+1-2x^2=0\)

\(\Leftrightarrow\)\(2x-3=0\)

\(\Leftrightarrow\)\(2x=3\)

\(\Leftrightarrow\)\(x=\frac{3}{2}\)

Vậy....

(x-2)(x+5)=x^2+5x-2x-10=x^2+3x-10

3x+3y+2x+2y=3(x+y)+2(x+y)=5(x+y)

(x^2+2xy):(x+2y)=x(x+2y):(x+2y)=x

x^2-4+x^2+2x+1-2x^2=0 khi 2x-3=0 khi x=3/2

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

(x^3 - 2x^2y + 3xy^2) : (-x)= -x^2+2xy-3y^2

Bạn hãy làm 1 điều coa thể

ok,tui nhất trí với bn nguyễn trọng bằng

\(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\)

\(=\left(x+\frac{1}{2}\right)\left(x-1\right)\left(x-\frac{2}{3}\right):\left(x+\frac{1}{2}\right)\)

\(=\left(x-1\right)\left(x-\frac{2}{3}\right)\)

a) Ta có: 6x^3 - 7x² - x+2 = 6x³+3x²-10x²-5x+4x+2

                                      = 3x² ( 2x+1) - 5x(2x+1) + 2(2x+1)

                                      = (2x+1)(3x²-5x+2)

Suy ra: (6x³-7x²-x+2): (2x+1)= 3x²-5x+2

b) Ta có: x² -y²+6x+9= (x²+6x+9) - y²

                               = (x+3)² - y²

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

Suy ra: (x²-y²+6x+9): (x+y+3)= x-y+3

Làm vậy nha pạn :) 

nếu tao biết tao chết liền

a)

\(=2x^2.5x^2−2x^2.2x+2x^2−3x.5x^2+3x.2x−3x\)

\(=10x^4−4x^3+2x^2−15x^3+6x^2−3x\)

\(=10x^4−19x^3+8x^2−3x\)

b) \((x−2y)(3xy+5y^2+x)\)

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

\(=3x^2y+5xy^2+x^2−6xy^2−10y^3−2xy\)

\(=3x^2y−xy^2−2xy+x^2−10y^3\)