\(\frac{-1^{100}}{16^{100}}=\frac{-1}{\left(2^4\right)^{100}}=\frac{-1}{2^{400}};\frac{-1^{500}}{2^{500}}=\frac{-1}{2^{500}}\)

Vì 2⁴⁰⁰<2⁵⁰⁰ => \(\frac{-1}{2^{400}}>\frac{-1}{2^{500}}\)=>\(\frac{-1^{100}}{16^{100}}>\frac{-1^{500}}{2^{500}}\)

Bài 1: \(\left(\frac{-1}{16}\right)^{100}=\frac{1}{\left(2^4\right)^{100}}=\frac{1}{2^{400}}>\frac{1}{2^{500}}=\left(\frac{-1}{2}\right)^{500}.\)

Bài 2: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)

\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)

Ta có:

\(\left(\dfrac{1}{16}\right)^{100}\)giữ nguyên

\(\left(\dfrac{-1}{2}\right)^{500}=\left[\left(\dfrac{-1}{2}\right)^5\right]^{100}=\left(\dfrac{-1}{32}\right)^{100}\)

Vì \(\left(\dfrac{1}{16}\right)^{100}>\left(\dfrac{-1}{32}\right)^{100}\Rightarrow\left(\dfrac{1}{16}\right)^{100}>\left(\dfrac{-1}{2}\right)^{500}\)

\(\frac{7x-5y}{500}=\frac{9x-5z}{300}=\frac{9y-7z}{100}=\frac{7xz-5yz}{500z}=\frac{9xy-5yz}{300y}=\frac{9xy-7zx}{100x}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{7x-5y}{500}=\frac{9x-5z}{300}=\frac{9y-7z}{100}=\)

\(\frac{7xz-5yz}{500z}=\frac{9xy-5yz}{300y}=\frac{9xy-7zx}{100x}=\frac{7xz-5yz-9xy+5yz+9xy-7zx}{500z-300y+100x}=0\)

\(\frac{7x-5y}{500}=0\Rightarrow7x=5y\Rightarrow\frac{x}{5}=\frac{y}{7}\)(1)

\(\frac{9x-5z}{300}=0\Rightarrow9x=5z\Rightarrow\frac{z}{9}=\frac{x}{5}\)(2)

\(\frac{9y-7z}{100}=0\Rightarrow9y=7z\Rightarrow\frac{y}{7}=\frac{z}{9}\)(3)

từ (1),(2),(3) => đpcm

a, Có: \(25^{200}=\left(5^2\right)^{200}=5^{400}\)

Vì \(5^{400}=5^{400}\) mà \(25^{200}=5^{400}\Rightarrow5^{400}=25^{200}\)

c, Có:

a/ 2⁶³ và 3⁴²

Ta có: 2⁶³=(2³)²¹=8²¹

3⁴²=(3²)²¹=9²¹

mà 8²¹<9²¹

vậy 2⁶³<3⁴²

b/5⁴⁰⁰ và 25²⁰⁰

Ta có: 25²⁰⁰=(5²)²⁰⁰=5⁴⁰⁰

mà 5⁴⁰⁰⁼5⁴⁰⁰

vậy 5⁴⁰⁰=25²⁰⁰

c/ \(\left(\dfrac{-1}{16}\right)^{100}v\text{à}\left(\dfrac{-1}{2}\right)^{500}\)

Ta có: \(\left(\dfrac{-1}{2}\right)^{500}=\left(\left(\dfrac{-1}{2}\right)^5\right)^{100}=\left(\dfrac{-1}{32}\right)^{100}\)

mà: \(\left(\dfrac{-1}{16}\right)^{100}< \left(\dfrac{-1}{32}\right)^{100}\)

vậy\(\left(\dfrac{-1}{16}\right)^{100}< \left(\dfrac{-1}{2}\right)^{500}\)

giúp mình đi làm ơn đó.mình cần rất gấp:) :) :)

mk lm rùi nên k cần giúp nx đâu

Bạn tham khảo nhé 

a )  Ta có : 

\(\left(-\frac{1}{5}\right)^{300}=\left(\frac{1}{5}\right)^{300}=\frac{1}{5^{300}}=\frac{1}{\left(5^3\right)^{100}}=\frac{1}{125^{100}}\)

\(\left(-\frac{1}{3}\right)^{500}=\left(\frac{1}{3}\right)^{500}=\frac{1}{3^{500}}=\frac{1}{\left(3^5\right)^{100}}=\frac{1}{243^{100}}\)

Do \(\frac{1}{125^{100}}>\frac{1}{243^{100}}\left(125^{100}< 243^{100}\right)\)

\(\Rightarrow\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)

b ) 

Ta có : 

\(2550^{10}=\left(50.51\right)^{10}=50^{10}.51^{10}\)

\(50^{20}=50^{10}.50^{10}\)

Do \(50^{10}.51^{10}>50^{10}.50^{10}\)

\(\Rightarrow50^{20}< 2550^{10}\)

c ) 

Ta có : 

\(2^{100}=\left(2^4\right)^{25}=16^{25}\)

\(3^{75}=\left(3^3\right)^{25}=27^{25}\)

\(5^{50}=\left(5^2\right)^{25}=25^{25}\)

Do \(16^{25}< 25^{25}< 27^{25}\)

\(\Rightarrow2^{100}< 5^{50}< 3^{75}\)

b)2550¹⁰>2500¹⁰=50²⁰

=>2550¹⁰>50²⁰