Toán $\frac{x+4}{x2-3x+2}$+ $\frac{x+1}{x2-4x+3}$+ $\frac{2x+5}{x2-4x+3}$ 27/10/2021 By Josie $\frac{x+4}{x2-3x+2}$+ $\frac{x+1}{x2-4x+3}$+ $\frac{2x+5}{x2-4x+3}$
ĐK: $x\notin\{1;2;3\}$ $\dfrac{x+4}{x^2-3x+2}+\dfrac{x+1+2x+5}{x^2-4x+3}$ $=\dfrac{x+4}{x^2-3x+2}+\dfrac{3x+6}{x^2-4x+3}$ $=\dfrac{x+4}{(x-1)(x-2)}+\dfrac{3x+6}{(x-1)(x-3)}$ $=\dfrac{(x+4)(x-3)+(3x+6)(x-2) }{(x-1)(x-2)(x-3)}$ $=\dfrac{x^2+x-12+3x^2-12 }{(x-1)(x-2)(x-3)}$ $=\dfrac{4x^2-x-24}{(x-1)(x-2)(x-3)}$ Trả lời
ĐK: $x\notin\{1;2;3\}$
$\dfrac{x+4}{x^2-3x+2}+\dfrac{x+1+2x+5}{x^2-4x+3}$
$=\dfrac{x+4}{x^2-3x+2}+\dfrac{3x+6}{x^2-4x+3}$
$=\dfrac{x+4}{(x-1)(x-2)}+\dfrac{3x+6}{(x-1)(x-3)}$
$=\dfrac{(x+4)(x-3)+(3x+6)(x-2) }{(x-1)(x-2)(x-3)}$
$=\dfrac{x^2+x-12+3x^2-12 }{(x-1)(x-2)(x-3)}$
$=\dfrac{4x^2-x-24}{(x-1)(x-2)(x-3)}$