a, (1-x)/(x+1 )+3=(2x+3)/(x+1)
b, ((x+2)^2)/(2x-3)-1=(x^2+10)/(2x-3)
c, (5x-2)/(2-2x)+(2x-1)/2=1- (x^2+x-3)/(1-x)
d, (5-2x)/3 + ((x-1)(x+1))/(3x-1)= ((x+2)(1-3x))/(9x-3)
a, (1-x)/(x+1 )+3=(2x+3)/(x+1)
b, ((x+2)^2)/(2x-3)-1=(x^2+10)/(2x-3)
c, (5x-2)/(2-2x)+(2x-1)/2=1- (x^2+x-3)/(1-x)
d, (5-2x)/3 + ((x-1)(x+1))/(3x-1)= ((x+2)(1-3x))/(9x-3)
a)$\frac{1-x}{x+1}$ +3=$\frac{2x+3}{x+1}$ ĐKXĐ:x $\neq$ -1
⇔$\frac{1-x}{x+1}$ +$\frac{3x+3}{x+1}$=$\frac{2x+3}{x+1}$
⇔1-x+3x+3=2x+3
⇔2x+4-2x-3=0
⇔1=0(vô lí)
Vậy S= rỗng
b)
$\frac{(x+2)^2}{2x-3}$-1= $\frac{x^2+10}{2x-3}$ ĐKXĐ: x $\neq$ $\frac{3}{2}$
⇔$\frac{x^2-2x+4}{2x-3}$-$\frac{2x-3}{2x-3}$= $\frac{x^2+10}{2x-3}$
⇔x^2-2x+4-2x+3=x^2+10
⇔x^2-4x+7-x^2-10=0
⇔-4x-3=0
⇔-4x=3
⇔x=-$\frac{3}{4}$ (TMĐK)
Vậy S ={-$\frac{3}{4}$ }
c)$\frac{5x-2}{2-2x}$ +$\frac{2x-1}{2}$= 1-$\frac{x^2+x-3}{1-x}$ ĐKXĐ:x$\neq$ 1
⇔$\frac{5x-2}{2(1-x)}$ +$\frac{(2x-1)(1-x)}{2(1-x)}$= $\frac{2(1-x)}{2(1-x)}$-$\frac{2(x^2+x-3)}{2(1-x)}$
⇔5x-2+(2x-1)(1-x)=2(1-x)-2(x^2+x-3)
⇔5x-2+2x-2x^2-1+x=2-2x-2x^2-2x+6
⇔8x-2x^2-3-2+2x+2x^2+2x-6=0
⇔12x-11=0
⇔x=$\frac{11}{12}$ (TMĐK)
Vậy S ={ $\frac{11}{12}$ }
d)$\frac{5-2x}{3}$+ $\frac{(x-1)(x+1)}{3x-1}$= $\frac{(x+2)(1-3x)}{9x-3}$ ĐKXĐ: x$\neq$ $\frac{1}{3}$
⇔$\frac{(5-2x)(3x-1)}{3(3x-1)}$+ $\frac{3(x-1)(x+1)}{3(3x-1)}$= $\frac{(x+2)(1-3x)}{3(3x-1)}$
⇔(5x-2)(3x-1)+3(x-1)(x+1)=(x+2)(1-3x)
⇔15x^2-5x-6x+2+3x^2-3=x-3x^2+2-6x
⇔18x^2-11x-1-x+3x^2-2+6x=0
⇔21x^2-6x-3=0
⇔3(7x^2-2x-1)=0
⇔7x^2-2x-1=0
⇔(đang nghĩ ;-; )