Bài 1

a: 11/12=1-1/12

23/24=1-1/24

mà -1/12>-1/24

nên 11/12>23/24

b: -3/20=-9/60

-7/12=-35/60

mà -9>-35

nên -3/20>-7/12

a) Ta có:

\(\dfrac{16}{9}\)=\(\dfrac{48}{27}\)                 \(\dfrac{24}{13}=\dfrac{48}{26}\)

Vì 27>26

➝\(\dfrac{48}{27}>\dfrac{48}{26}hay\dfrac{16}{9}>\dfrac{24}{13}\)

So sánh:

a) 16/9 và 24/13 

 Ta có \(\dfrac{16}{9}=\dfrac{208}{117}\) và \(\dfrac{24}{13}=\) \(\dfrac{216}{117}\)

\(\Rightarrow\dfrac{216}{117}>\dfrac{208}{117}\Rightarrow\dfrac{24}{13}>\dfrac{16}{9}\)

b) 27/82 và 26/75

Ta có \(\dfrac{27}{82}\approx0,33\) và \(\dfrac{26}{75}\approx0,35\)

\(\Rightarrow9,35>0,33\Rightarrow\dfrac{26}{75}>\dfrac{27}{82}\)

a) Ta có: \(\dfrac{16}{9}=\dfrac{16\cdot13}{9\cdot13}=\dfrac{208}{117}\)

\(\dfrac{24}{13}=\dfrac{24\cdot9}{13\cdot9}=\dfrac{216}{117}\)

mà \(\dfrac{208}{117}< \dfrac{216}{117}\)

nên \(\dfrac{16}{9}< \dfrac{24}{13}\)

a) Vì \(\dfrac{1}{24}< \dfrac{1}{83}\) 

⇒ \(\dfrac{1}{24^9}>\dfrac{1}{83^{13}}\)

a) \(\left(\dfrac{1}{24}\right)^9>\left(\dfrac{1}{27}\right)^9=\dfrac{1}{3^{27}}\)

\(\left(\dfrac{1}{83}\right)^{13}< \left(\dfrac{1}{81}\right)^{13}=\dfrac{1}{3^{52}}\)

Mà \(\dfrac{1}{3^{27}}>\dfrac{1}{3^{52}}\)

\(\Rightarrow\left(\dfrac{1}{24}\right)^9>\left(\dfrac{1}{83}\right)^{13}\)

b) \(3^{300}=\left(3^3\right)^{100}=27^{100}\)

\(5^{199}< 5^{200}=\left(5^2\right)^{100}=25^{100}\)

Mà \(25^{100}< 27^{100}\)

\(\Rightarrow5^{199}< 3^{300}\)

\(\Rightarrow\dfrac{1}{5^{199}}>\dfrac{1}{3^{300}}\)

\(a,\left(\dfrac{1}{24}\right)^9=\dfrac{1}{24^9};\left(\dfrac{1}{83}\right)^{13}=\dfrac{1}{83^{13}};24^9< 83^{13}\left(24< 83;9< 13\right)\\ \Rightarrow\dfrac{1}{24^9}< \dfrac{1}{83^{13}}\Rightarrow\left(\dfrac{1}{24}\right)^9< \left(\dfrac{1}{83}\right)^{13}\\ b,3^{300}=27^{100}>25^{100}=5^{200}>5^{199}\\ \Rightarrow\dfrac{1}{3^{300}}< \dfrac{1}{5^{199}}\)

\(9^{12}=\left(3^2\right)^{12}=3^{24}=\left(3^3\right)^8=27^8>26^8\)

Có 9^12=(3^2)^12=3^24=3^3.8=(3^3)^8=27^8

Ma 27^8>36^8 suy ra 9^12>26^8

Vay 9^12>26^8

Bài 6: 

a: \(15=\sqrt{225}>\sqrt{200}\)

b: \(27=9\sqrt{9}>9\sqrt{5}\)

c: \(-24=-\sqrt{576}< -\sqrt{540}=-6\sqrt{15}\)

 \(a)\) Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{\text{ }10}\)

Vì \(9801< 9999\)nên \(9801^{10}< 9999^{10}\)

Hay \(99^{20}< 9999^{10}\)

Vậy 99²⁰<9999¹⁰