a: \(\dfrac{2x^4-x^3-x^2+7x-4}{x^2+x-1}\)

\(=\dfrac{2x^4+2x^3-2x^2-3x^3-3x^2+3x+4x^2+4x-4}{x^2+x-1}\)

=2x^2-3x+4

b: \(=\dfrac{y}{x\left(2x-y\right)}+\dfrac{4x}{y\left(y-2x\right)}\)

\(=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(2x-y\right)\left(2x+y\right)}{xy\left(2x-y\right)}=\dfrac{-2x-y}{xy}\)

c: \(=\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)

`(x^2-y^2)/(x+y)`

`=((x-y)(x+y))/(x+y)`

`=x-y`

\(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)

\(a.2x^3+6x=2x\left(x^2+3\right)\)

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

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

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

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

\(c.3x\left(x-5y\right)-2y\left(5y-x\right)\)

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

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

\(d.y^2\left(x^2+y\right)-x^3-xy\)

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

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

e. Cái bài này ghi lại đàng hoàng xíu nha t k hỉu

\(f.3x^2\left(y^2-2x\right)-15x\left(2x-y^2\right)\)

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

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

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)

e.

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

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

f.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

Đề bài yêu cầu gì vậy em.

a, (x^2 -2x+1)+(y^2 +6y+9) =0

(x-1)^2 +(y+3)^2 =0

Do đó: x-1=0 và y+3=0

Vậy x=1 và y=-3

b, x^2 +y^2 +1=xy+x+y

2x^2 +2y^2 +2=2xy+2x+2y

2x^2 +2y^2 -2xy-2x-2y +2=0

(x^2 -2x+1)+(y^2 -2y+1)+ (x^2 +y^2 -2xy)=0

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

Suy ra: x-1=0, y-1=0 và x-y=0

Vậy x=1,y=1

c,5x^2 - 4x-2xy+y^2 +1=0

(4x^2 -4x+1)+(x^2 -2xy+y^2 )=0

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

Do đó: 2x-1 =0 và x=y suy ra: x=0,5 và x=y

Vậy x=y=0,5

a: \(=\left(3-x\right)\left(x+1\right)\)

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

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

c: \(=x\left(x-y\right)-10\left(x-y\right)\)

\(=\left(x-y\right)\left(x-10\right)\)

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

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

c) \(=x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(x-10\right)\)

d) \(=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)

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

f) \(=9-\left(4x^2-4xy+y^2\right)=9-\left(2x-y\right)^2=\left(3-2x+y\right)\left(3+2x-y\right)\)

g) \(=y\left(y^2-2xy+x^2-y\right)\)

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

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

\(a,=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\\ b,=\left(x+y\right)\left(x-5\right)\\ c,=5x^2\left(x-y\right)-10x\left(x-y\right)=5x\left(x-2y\right)\left(x-y\right)\\ d,=x^2-2xy=x\left(x-2y\right)\\ e,=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)