Ta có :
\(\frac{-1}{2}^{300}=\left[\left(-\frac{1}{2}\right)^3\right]^{100}=\left(-\frac{1}{8}\right)^{100}\)
\(\frac{-1}{3}^{200}=\left[\left(-\frac{1}{3}\right)^2\right]^{100}=\frac{1}{9}^{100}\)
vì \(\left(-\frac{1}{8}\right)^{100}=\frac{1}{8}^{100}\)mà 8100 < 9100 nên \(\frac{1}{8}^{100}>\frac{1}{9}^{100}\)hay \(\left(-\frac{1}{8}\right)^{100}>\left(\frac{1}{9}\right)^{100}\)
Vậy \(\left(-\frac{1}{2}\right)^{300}>\left(-\frac{1}{3}\right)^{200}\)
\(\left(\frac{-1}{2}\right)^{300}=\left[\left(\frac{-1}{2}\right)^3\right]^{100}=\left(\frac{-1}{8}\right)^{100}\)
\(\left(\frac{-1}{3}\right)^{200}=\left[\left(\frac{-1}{3}\right)^2\right]^{100}=\left(\frac{1}{9}\right)^{100}\)
vì \(\left(\frac{-1}{8}\right)^{100}< \left(\frac{1}{9}\right)^{100}\)nên \(\left(\frac{-1}{2}\right)^{300}< \left(\frac{-1}{3}\right)^{200}\)
\(\frac{-1^{300}}{2}=\frac{-1}{2}\)
\(\frac{-1^{200}}{3}=\frac{-1}{3}\)
Vi \(2< 3\) nen \(\frac{-1}{2}< \frac{-1}{3}\)
\(\Rightarrow\frac{-1^{300}}{2}< \frac{-1^{200}}{3}\)
Nhìn trông thì đáng sợ nhưng đơn giản lắm bạn
Ta có \(\frac{-1^{300}}{2}=\frac{-1}{2}\)
\(\frac{-1^{200}}{3}=\frac{-1}{3}\)
Vì \(\frac{-1}{2}< \frac{-1}{3}\Rightarrow\frac{-1^{300}}{2}< \frac{-1^{200}}{3}\)
Vậy \(\frac{-1^{300}}{2}< \frac{-1^{200}}{3}\)
P/s : vì -1 ko có ngoặc nên chỉ tính trên tử thôi nhé