Ta có \(\frac{32x-8x^2+2x^3}{x^3+64}=\frac{x\left(32-8x+2x^2\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{2x}{x+4}\)

\(\dfrac{x^3+64}{2x^3-8x^2+32x}\\ =\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}\\ =\dfrac{x+4}{2x}\)

\(\dfrac{x^3+64}{2x^3-8x^3+32x}\)

\(=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}\)

\(=\dfrac{x+4}{2x}\)

a) (2x^2 +2xy - xy -y^2 ) / (2x^2 - 2xy - xy +y^2)

= 2x(x+y) - y(x+y)  /  2x(x-y) - y(x-y)

= (2x-y)(x+y)  /  (2x-y)(x-y)

= x+y/x-y

Rút gọn cái sau:

\(\frac{32x+4x^2+2x^3}{x^3+64}\)

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

Đề có vẻ sai sai ? 

\(\frac{x}{8}=\frac{-15}{30}\)

\(\Rightarrow x=\frac{-15.8}{30}\)

\(\Rightarrow x=-4\)

\(\frac{y}{6}=\frac{15}{-45}\)

\(\Rightarrow y=\frac{15.6}{-45}\)

\(\Rightarrow y=-2\)

A)           \(\frac{x}{8}=\frac{-15}{30}\)

\(\Leftrightarrow\)\(x=\frac{\left(-15\right).8}{30}=-4\)

Vậy...

B)    \(\frac{y}{6}=\frac{15}{-45}\)

\(\Leftrightarrow\)\(y=\frac{15.6}{-45}=-2\)

Vậy....

a/b=c/b => a/c=b/d, d/b=c/a, d/c=b/a

ta co 16 = 2⁴

         24 = 2³. 3

         56 = 2³.7

=> BCNN(16,24,56)=2⁴.3.7=2⁴.3.7=336

ta co: \(\frac{-3}{16}=\frac{-63}{336}\)

         \(\frac{5}{24}=\frac{70}{336}\)

           \(\frac{-21}{56}=\frac{-126}{336}\)