Question

Giúp em vs ạ

Toán đại bài phân thức đại số ạ

a, x^2y^3/5 = 7x^3y^4/ 35xy

b, x^3-4x/10-5x=-x^2-2x/5

c,x+2/x-1=(x+2)(x+1)/x^2-1

d, x^2-x-2/x+1=x^2-3x+2/x-1

e, x^3+8/x^2-2x+4=x+2

(Giúp em vs ạ ,em cảm ơn trc nha)

Answer 1

a: \(\dfrac{7x^3y^4}{35xy}=\dfrac{7xy\cdot x^2y^3}{7xy\cdot5}=\dfrac{x^2y^3}{5}\)

b: \(\dfrac{x^3-4x}{10-5x}=\dfrac{x\left(x-2\right)\left(x+2\right)}{-5\left(x-2\right)}=\dfrac{-x\left(x+2\right)}{5}=\dfrac{-x^2-2x}{5}\)

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

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

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

=(x-2)(x^2-1)

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

e: \(\dfrac{x^3+8}{x^2-2x+4}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{x^2-2x+4}=x+2\)