Question

quy đồng mẫu phân thức sau

\(\dfrac{4x^2-3x+5}{x^3-1}\) , \(\dfrac{1-2x}{x^2+x+1}\) và -2

Answer 1

\(\dfrac{4x^2-3x+5}{x^3-1}=\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

\(\dfrac{4x^2-3x+5}{x^3-1}=\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

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

Answer 2

ta có:MTC=x³-1

Quy đồng:

\(\dfrac{4x^2-3x+5}{x^3-1}\)

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

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