Question

[thùy linh]

bài 11.rút gọn biểu thức:

\(a,\dfrac{9x^2}{11y^2}:\dfrac{3x}{2y}:\dfrac{6x}{11y}\)              \(b,\dfrac{3x+15y}{x^3-y^3}:\dfrac{x+5y}{x-y}\)         

 \(c,\dfrac{x^2-1}{x^2-4x+4}:\dfrac{x+1}{2-x}\)          \(d,\dfrac{5x+10}{x+2}:\dfrac{5y}{x}\)

         \(e,\dfrac{2x}{3x-3y}:\dfrac{x^2}{x-y}\)          \(f,\dfrac{5x-3}{4x^2y}-\dfrac{x-3}{4x^2y}\)       

  \(g,\dfrac{3x+10}{x+3}-\dfrac{x+4}{x+3}\)                \(h,\dfrac{4}{x-1}+\dfrac{2}{1-x}+\dfrac{x}{x-1}\)

       \(i,\dfrac{2x^2-x}{x-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}\)     \(j,\dfrac{x-2}{x-6}-\dfrac{x-18}{6-x}+\dfrac{x+2}{x-6}\)

\(k,\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\)       \(m,\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)       

 \(n,\dfrac{3}{x+3}-\dfrac{x-6}{x^2+3x}\)              \(p,\dfrac{x+3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)

g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)

h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)

n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)

p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)

k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)

m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)