Rút gọn phân thức: `(3x^2-2x-1)/(3x^3+4x^2-5x-2)` 27/11/2021 Bởi Reese Rút gọn phân thức: `(3x^2-2x-1)/(3x^3+4x^2-5x-2)`
Đáp án: $\dfrac{1}{x+2}$ Giải thích các bước giải: Ta có: $\dfrac{3x^2-2x-1}{3x^3+4x^2-5x-2}$ $=\dfrac{(3x^2-3x)+(x-1)}{(3x^3-3x^2)+(7x^2-7x)+(2x-2)}$ $=\dfrac{3x(x-1)+(x-1)}{3x^2(x-1)+7x(x-1)+2(x-1)}$ $=\dfrac{(3x+1)(x-1)}{(x-1)(3x^2+7x+2)}$ $=\dfrac{(3x+1)(x-1)}{(x-1)((3x^2+6x)+(x+2))}$ $=\dfrac{(3x+1)(x-1)}{(x-1)(3x(x+2)+(x+2))}$ $=\dfrac{(3x+1)(x-1)}{(x-1)(3x+1)(x+2)}$ $=\dfrac{1}{x+2}$ Bình luận
Đáp án: $\dfrac{1}{x+2}$
Giải thích các bước giải:
Ta có:
$\dfrac{3x^2-2x-1}{3x^3+4x^2-5x-2}$
$=\dfrac{(3x^2-3x)+(x-1)}{(3x^3-3x^2)+(7x^2-7x)+(2x-2)}$
$=\dfrac{3x(x-1)+(x-1)}{3x^2(x-1)+7x(x-1)+2(x-1)}$
$=\dfrac{(3x+1)(x-1)}{(x-1)(3x^2+7x+2)}$
$=\dfrac{(3x+1)(x-1)}{(x-1)((3x^2+6x)+(x+2))}$
$=\dfrac{(3x+1)(x-1)}{(x-1)(3x(x+2)+(x+2))}$
$=\dfrac{(3x+1)(x-1)}{(x-1)(3x+1)(x+2)}$
$=\dfrac{1}{x+2}$