Giải PT: $\frac{18}{x^2 +4x+3}$ +$\frac{18}{(x+3)(x+5)}$ = $x^{2}$ +6x -4 Cần trc 6h!! ;; 30/09/2021 Bởi Madeline Giải PT: $\frac{18}{x^2 +4x+3}$ +$\frac{18}{(x+3)(x+5)}$ = $x^{2}$ +6x -4 Cần trc 6h!! ;;
Đáp án: $ x\in\{1, -7, -2, -4\}$ Giải thích các bước giải: ĐKXĐ: $x\ne -1, -3 , -5$ Ta có: $\dfrac{18}{x^2+4x+3}+\dfrac{18}{(x+3)(x+5)}=x^2+6x-4$ $\to \dfrac{18}{(x+1)(x+3)}+\dfrac{18}{(x+3)(x+5)}=x^2+6x-4$ $\to \dfrac{9\cdot 2}{(x+1)(x+3)}+\dfrac{9\cdot 2}{(x+3)(x+5)}=x^2+6x-4$ $\to 9\cdot (\dfrac{2}{(x+1)(x+3)}+\dfrac{2}{(x+3)(x+5)})=x^2+6x-4$ $\to 9\cdot (\dfrac{(x+3)-(x+1)}{(x+1)(x+3)}+\dfrac{(x+5)-(x+3)}{(x+3)(x+5)})=x^2+6x-4$ $\to 9\cdot (\dfrac1{x+1}-\dfrac1{x+3}+\dfrac1{x+3}-\dfrac1{x+5})=x^2+6x-4$ $\to 9\cdot (\dfrac1{x+1}-\dfrac1{x+5})=x^2+6x-4$ $\to 9\cdot \dfrac{4}{(x+1)(x+5)}=x^2+6x-4$ $\to \dfrac{36}{x^2+6x+5}=x^2+6x-4$ Đặt $x^2+6x+5=t$ $\to \dfrac{36}{t}=t-9$ $\to t(t-9)=36$ $\to t\in\{12, -3\}$ $\to x^2+6x+5\in\{12, -3\}$ $\to x\in\{1, -7, -2, -4\}$ Bình luận
Đáp án: $ x\in\{1, -7, -2, -4\}$
Giải thích các bước giải:
ĐKXĐ: $x\ne -1, -3 , -5$
Ta có:
$\dfrac{18}{x^2+4x+3}+\dfrac{18}{(x+3)(x+5)}=x^2+6x-4$
$\to \dfrac{18}{(x+1)(x+3)}+\dfrac{18}{(x+3)(x+5)}=x^2+6x-4$
$\to \dfrac{9\cdot 2}{(x+1)(x+3)}+\dfrac{9\cdot 2}{(x+3)(x+5)}=x^2+6x-4$
$\to 9\cdot (\dfrac{2}{(x+1)(x+3)}+\dfrac{2}{(x+3)(x+5)})=x^2+6x-4$
$\to 9\cdot (\dfrac{(x+3)-(x+1)}{(x+1)(x+3)}+\dfrac{(x+5)-(x+3)}{(x+3)(x+5)})=x^2+6x-4$
$\to 9\cdot (\dfrac1{x+1}-\dfrac1{x+3}+\dfrac1{x+3}-\dfrac1{x+5})=x^2+6x-4$
$\to 9\cdot (\dfrac1{x+1}-\dfrac1{x+5})=x^2+6x-4$
$\to 9\cdot \dfrac{4}{(x+1)(x+5)}=x^2+6x-4$
$\to \dfrac{36}{x^2+6x+5}=x^2+6x-4$
Đặt $x^2+6x+5=t$
$\to \dfrac{36}{t}=t-9$
$\to t(t-9)=36$
$\to t\in\{12, -3\}$
$\to x^2+6x+5\in\{12, -3\}$
$\to x\in\{1, -7, -2, -4\}$