chứng minh hai phân thức bằng nhau a)2x^4+3x^3+2x+3/(x^2-x-1)(4x+6)=x+1/2 27/08/2021 Bởi Hadley chứng minh hai phân thức bằng nhau a)2x^4+3x^3+2x+3/(x^2-x-1)(4x+6)=x+1/2
Ta có $\dfrac{2x^4 + 3x^3 + 2x + 3}{(x^2-x+1)(4x+6)} = \dfrac{x^3(2x+3) + (2x+3)}{(x^2-x+1)(4x+6)}$ $= \dfrac{(x^3+1)(2x+3)}{(x^2-x+1).2(2x+3)}$ $= \dfrac{(x+1)(x^2-x+1)}{2(x^2-x+1)}$ $= \dfrac{x+1}{2} = VP$ Bình luận
Ta có
$\dfrac{2x^4 + 3x^3 + 2x + 3}{(x^2-x+1)(4x+6)} = \dfrac{x^3(2x+3) + (2x+3)}{(x^2-x+1)(4x+6)}$
$= \dfrac{(x^3+1)(2x+3)}{(x^2-x+1).2(2x+3)}$
$= \dfrac{(x+1)(x^2-x+1)}{2(x^2-x+1)}$
$= \dfrac{x+1}{2} = VP$