a)

\(\dfrac{-2}{3}\)>\(\dfrac{5}{-8}\)

b)

\(\dfrac{398}{-412}\)<\(\dfrac{-25}{-137}\)

c)

\(\dfrac{-14}{21}\)<\(\dfrac{60}{72}\)

a <

b <

c <

a)

5/−8

b)

−25/−137

c)

60/72

a: B=31/39=62/78=1-16/78; A=35/37=280/296=1-16/296

16/78>16/296

=>B<A

b: C=23/27=46/54=1-8/54

D=1-8/29

8/29>8/54

=>D<C

a) `14/21=(14:7)/(21:7)=2/3=4/6`

`60/72=(60:12)/(72:12)=5/6`

Vì `4/6 <5/6`

`=> 14/21 < 60/72`

b) `22/37 = (22:2)/(37:2)= 11/(37/2)`

Vì `54 > 37/2`

`=> 11/54 < 22/37` 

help 

Bài 21: 

Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\Leftrightarrow\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1+\dfrac{x+3}{97}+1+\dfrac{x+4}{96}+1=0\)

\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)

mà \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}>0\)

nên x+100=0

hay x=-100

Vậy: x=-100

a)2^6=8^2

b)5^3<3^5

a) \(2^6\) và \(8^2\)

   \(2^6=\left(2^2\right)^3\)

  \(8^2=\left(2^3\right)^2\)\(=2^6\)

 \(\Rightarrow\) \(2^6=8^2\)

bài nào

bài nào bn

???

a: 99^20=9801^10<9999^10

b: 3^500=243^100

5^300=125^300

=>3^500>5^300

a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!