Ta có:

\(\left(\frac{-1}{8}\right)^{100}=\frac{\left(-1\right)^{100}}{8^{100}}=\frac{1}{\left(2^3\right)^{100}}=\frac{1}{2^{300}}\)

\(\left(\frac{-1}{4}\right)^{200}=\frac{\left(-1\right)^{200}}{4^{200}}=\frac{1}{\left(2^2\right)^{100}}=\frac{1}{2^{200}}\)

Vì \(2^{300}>2^{200}\)\(\Rightarrow\frac{1}{2^{300}}< \frac{1}{2^{200}}\)

\(\Rightarrow\left(\frac{-1}{8}\right)^{^{100}}< \left(\frac{-1}{4}\right)^{200}\)

bằng nhau

ta có:\(\left(-\frac{1}{8}\right)^{180}=\left(\frac{1}{8}\right)^{180}=\left(\frac{1}{4}\right)^{2^{180}}=\left(\frac{1}{4}\right)^{360}\) 

ta có :\(\left(-\frac{1}{4}\right)^{200}=\left(\frac{1}{4}\right)^{200}\)

=>(1/4)^360<(1/4)^200

Vậy : (-1/8)^180 < ( -1/4)^200

Ta có: \(\left(\frac{-1}{8}\right)^{180}=\frac{-1^{180}}{8^{180}}=\frac{1}{8^{180}}\)

\(\left(\frac{-1}{4}\right)^{200}=\frac{-1^{200}}{4^{200}}=\frac{1}{4^{200}}\)

Suy ra để so sánh \(\left(\frac{-1}{8}\right)^{180}\)và\(\left(\frac{-1}{4}\right)^{200}\)thì ta chỉ cần so sánh 8¹⁸⁰ và 4²⁰⁰

Ta có: 8¹⁸⁰=(2³)¹⁸⁰=2^3.180=2⁵⁴⁰

             4²⁰⁰=(2²)²⁰⁰=2^2.200=2⁴⁰⁰

       Ta thấy 2=2 nhưng 540>400 suy ra 8¹⁸⁰>4²⁰⁰ suy ra \(\frac{1}{8^{180}}< \frac{1}{4^{200}}\)suy ra \(\left(\frac{-1}{8}\right)^{180}< \left(\frac{-1}{4}\right)^{200}\)

ta có:1/8^100

       -1/4^200=(-1/4^2)^100=1/16^100

=>1/8^100 >1/16^100

=>1/8^100 >-1/4^200

\(\left(\dfrac{1}{16}\right)^{200}< \left(\dfrac{1}{2}\right)^{1000}\)

16 = 2⁴

(\(\dfrac{1}{16}\))²⁰⁰ = \(\dfrac{1}{2^{4.200}}\) = \(\dfrac{1}{2^{800}}\)= (\(\dfrac{1}{2}\))⁸⁰⁰

So sánh với (\(\dfrac{1}{2}\))¹⁰⁰⁰

Hai phân số cùng tử số, phân số nào có mẫu lớn hơn thì phân số đó nhỏ hơn

Suy ra: (\(\dfrac{1}{16}\))²⁰⁰ > (\(\dfrac{1}{2}\))¹⁰⁰⁰

Ta có: \(\left(\dfrac{1}{16}\right)^{200}=\left(\dfrac{1}{2}\right)^{800}\)

mà \(\left(\dfrac{1}{2}\right)^{800}>\left(\dfrac{1}{2}\right)^{1000}\)

nên \(\left(\dfrac{1}{16}\right)^{200}< \left(\dfrac{1}{2}\right)^{1000}\)

làm được bài 1:

TA CÓ: \(\left(\frac{1}{16}\right)^{200}=\left(\frac{1}{16}\right)^{200}\)

            \(\left(\frac{1}{2}\right)^{1000}=\left(\frac{1}{2}\right)^{5.200}=\left(\frac{1^5}{2^5}\right)^{200}=\left(\frac{1}{32}\right)^{200}\)

vì mũ số bằng nhau nên ta so sánh phân số. Vì \(\frac{1}{16}>\frac{1}{32}\)nên \(\left(\frac{1}{16}\right)^{200}>\left(\frac{1}{32}\right)^{200}\)do đó\(\left(\frac{1}{16}\right)^{200}>\left(\frac{1}{2}\right)^{1000}\)

Cho mk sửa xíu : ( 1/2)^-1 ; (1/2)^-2

(1/2)^-1=2

(1/2)^-2=4

có 2<4

=>(1/2)^-1<(1/2)^-2

Ta có :

 \(\left(\frac{1}{2}\right)^{-1}=\left(2^{-1}\right)^{-1}=2\)

\(\left(\frac{1}{3}\right)^{-1}=3\)

\(\left(\frac{1}{2}\right)^{-2}=\left(2^{-1}\right)^{-2}=2^2=4\)

\(\left(\frac{1}{4}\right)^{-1}=\left(4^{-1}\right)^{-1}=4\)

\(\left(\frac{1}{3}\right)^{-2}=\left(3^{-1}\right)^{-2}=3^2=9\)

Do đó ta có :

\(\left(\frac{1}{2}\right)^{-1}< \left(\frac{1}{3}\right)^{-1}< \left(\frac{1}{2}\right)^{-2}=\left(\frac{1}{4}\right)^{-1}< \left(\frac{1}{3}\right)^{-2}\)