a ) [tex]\frac{1}{2x+4033}+\frac{44}{x+2000}=-3[/tex]
b ) [tex]\frac{1}{6x+5}+\frac{4}{3x+1}+\frac{-3}{2x+3}=-6[/tex]
a) Đề đúng ko bạn :v
1. $\dfrac{1}{2x+40\color{red}{4}3}+\dfrac{44}{x+2000}=-3$
ĐK: $x\ne -2000;x\ne \dfrac{-4043}2$
pt $\Leftrightarrow \dfrac{1}{2x+4043}+1+\dfrac{44}{x+2000}+2=0$
$\Leftrightarrow \dfrac{2x+4044}{2x+4033}+\dfrac{2x+4044}{x+2000}=0$
$\Leftrightarrow (2x+4044)(\dfrac{1}{2x+4043}+\dfrac{1}{x+2000})=0$
$\Leftrightarrow 2(x+2022).\dfrac{3x+6043}{(2x+4033)(x+2000)}=0$
$\Leftrightarrow x=-2022$ or $x=\dfrac{-6043}3$ (TM)
Vậy
2. $\dfrac{1}{2x+4033}+\dfrac{\color{red}{3}4}{x+2000}=-3$
ĐK: $x\ne -2000;x\ne \dfrac{-4033}2$
pt $\Leftrightarrow \dfrac{1}{2x+4043}+1+\dfrac{34}{x+2000}+2=0$
$\Leftrightarrow \dfrac{2x+4034}{2x+4033}+\dfrac{2x+4034}{x+2000}=0$
$\Leftrightarrow (2x+4034)(\dfrac{1}{2x+4033}+\dfrac{1}{x+2000})=0$
$\Leftrightarrow 2(x+2017).\dfrac{3x+6033}{(2x+4033)(x+2000)}=0$
$\Leftrightarrow x=-2017$ or $x=-2011$ (TM)
Vậy...
b) ĐK: $x\ne \dfrac{-5}6;x\ne \dfrac{-1}3;x\ne \dfrac{-3}2$
pt $\Leftrightarrow \dfrac{1}{6x+5}+1+\dfrac{4}{3x+1}+2+\dfrac{-3}{2x+3}+3=0$
$\Leftrightarrow \dfrac{6x+6}{6x+5}+\dfrac{6x+6}{3x+1}+\dfrac{6x+6}{2x+3}=0$
$\Leftrightarrow 6(x+1)(\dfrac{1}{6x+5}+\dfrac{1}{3x+1}+\dfrac{1}{2x+3})=0$
$\Leftrightarrow x=-1$ or $\dfrac{1}{6x+5}+\dfrac{1}{3x+1}+\dfrac{1}{2x+3}=0 \ (*)$
pt $(*)\Leftrightarrow \dfrac{(3x+1)(2x+3)+(6x+5)(2x+3)+(6x+5)(3x+1)}{(6x+5)(3x+1)(2x+3)}=0$
$\Leftrightarrow \dfrac{36x^2+60x+23}{(6x+5)(3x+1)(2x+3)}=0$
$\Rightarrow 36x^2+60x+23=0$
$\Leftrightarrow 36x^2+60x+25-2=0$
$\Leftrightarrow (6x+5)^2=2$
$\Leftrightarrow 6x+5=\pm \sqrt 2$
$\Leftrightarrow x=\dfrac{-5\pm \sqrt 2}6$ (TM)
Vậy...
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
