a/ $\dfrac{x+5}{7}+\dfrac{x+3}{9}=\dfrac{x+1}{11}+\dfrac{x-1}{13}\\↔\left(\dfrac{x+5}{7}+1\right)+\left(\dfrac{x+3}{9}+1\right)=\left(\dfrac{x+1}{11}+1\right)+\left(\dfrac{x-1}{13}+1\right)\\↔\dfrac{x+12}{7}+\dfrac{x+12}{9}=\dfrac{x+12}{11}+\dfrac{x+12}{13}\\↔\dfrac{x+12}{7}+\dfrac{x+12}{9}-\dfrac{x+12}{11}-\dfrac{x+12}{13}=0\\↔(x+12)\left(\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}-\dfrac{1}{13}\right)=0$
Ta có: $\dfrac{1}{7}>\dfrac{1}{11},\,\dfrac{1}{9}>\dfrac{1}{13}$
$→\dfrac{1}{7}+\dfrac{1}{9}>\dfrac{1}{11}+\dfrac{1}{13}\\↔\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}-\dfrac{1}{13}>0\\→\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}-\dfrac{1}{13}\ne 0$
mà $(x+12)\left(\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}-\dfrac{1}{13}\right)=0$
$→x+12=0\\↔x=-12$
Vậy pt có tập nghiệm $S=\{-12\}$
b/ ĐKXĐ: $x\in\Bbb R$
$\sqrt{x^2+x+1}=3-x\\↔x^2+x+1=(3-x)^2\\↔x^2+x+1=9-6x+x^2\\↔7x=8\\↔x=\dfrac{8}{7}$
Vậy pt có tập nghiệm $S=\left\{\dfrac{8}{7}\right\}$
