a) \(\frac{31}{91}\) và \(\frac{311}{911}\)
Ta có: \(\frac{31}{91}=\frac{31\cdot10}{91\cdot10}=\frac{310}{910}\)
Áp dụng \(\frac{a}{b}< \frac{a+m}{b+m}\) (a,b ϵ Z ; b \(\ne\) 0 ; a<b)
⇒ \(\frac{310}{910}< \frac{310+1}{910+1}=\frac{311}{911}\)
⇒ \(\frac{31}{91}< \frac{311}{911}\)
b) \(\left(\frac{1}{16}\right)^{30}\) và \(\left(\frac{-1}{8}\right)^{50}\)
⇒ \(\left(\frac{1}{16}\right)^{30}=\left(\left(\frac{1}{2}\right)^4\right)^{30}=\left(\frac{1}{2}\right)^{120}\) (1)
⇒ \(\left(\frac{-1}{8}\right)^{50}=\left(\left(\frac{-1}{2}\right)^3\right)^{50}=\left(\frac{-1}{2}\right)^{150}\)
mà \(\left(-a\right)^m=a^m\) (m chẵn)
⇒\(\left(\frac{-1}{8}\right)^{50}=\left(\frac{-1}{2}\right)^{150}=\left(\frac{1}{2}\right)^{150}\) (2)
Từ (1) và (2) ⇒ \(\left(\frac{1}{2}\right)^{120}>\left(\frac{1}{2}\right)^{150}\) hay \(\left(\frac{1}{16}\right)^{30}>\left(\frac{-1}{8}\right)^{50}\)
Câu C dài quá :))