Question

[Thu Phương]

Rút gọn các phân thức sau :
a) \(\dfrac{x^2-16 }{4x-x^2}\) ( x \(\ne\) x , x \(\ne\) 4 )
b) \(\dfrac{x^2+4x+3}{2x+6}\) ( x \(\ne\) -3 )

c) \(\dfrac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}\) ( y + ( x + y ) \(\ne\) 0 )

d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}\) ( x \(\ne\) y )

e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}\) ( x \(\ne\) - y )

f)\(\dfrac{x^2-xy}{3xy-3y^2}\) ( x \(\ne\) y , y \(\ne\) 0 )

g) \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\) ( b \(\ne\) 0 , x \(\ne\pm\)1 )

h) \(\dfrac{4x^2-4xy}{5x^3-5x^2y}\left(x\ne0,x\ne y\right)\)

i) \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\left(x+y+z\ne0\right)\)

k)\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\left(x\ne0,x\ne y\right)\)
Help me!!!

Answer 1

[Akai Haruma]

a)

\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)

c)

\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)

d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)

Answer 2

[Akai Haruma]

e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)

f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)

g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)

\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)