Question

biến đổi các biểu thức sau thành phân thức

a, \(\dfrac{1}{2}+\dfrac{x}{1-\dfrac{x}{x+2}}\)

b,\(\dfrac{x-\dfrac{1}{x^2}}{1+\dfrac{1}{x}+\dfrac{1}{x^2}}\)

Answer 1

a) \(\dfrac{1}{2}+\left[x:\left(1-\dfrac{x}{x+2}\right)\right]=\dfrac{1}{2}+\left(x:\dfrac{x+2-x}{x+2}\right)\)

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

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

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