[tex]\frac{3(x-1)}{x-1} + \frac{10}{x-1}=3+\frac{10}{x-1} <=> \frac{3x-3}{x-1} + \frac{10}{x-1} - \frac{10}{x-1}=3 <=> 3x-3 = 3x-3(tu giai nhe) <=> x\epsilon R => x\epsilon R, x\neq 1[/tex]
[tex]\frac{4x-12}{3-x}+\frac{11}{3-x}= \frac{4x-12+11}{3-x}= \frac{4x-1}{3-x} x\neq 3[/tex]
Mình đoán vậy!!!