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

(0.3)⁴⁰=((0.3)²)²⁰=(0.09)²⁰   Do 0.1>0.09  =>(0.1)²⁰ > (0.09)²⁰   <=> (0.1)²⁰  > (0.3)⁴⁰

(-5)³⁰=((-5)³)¹⁰=(-125)¹⁰ =125¹⁰        (-3)⁵⁰=((-3)⁵)¹⁰=(-243)¹⁰ =243¹⁰      Do 125<243   =>125¹⁰ < 243¹⁰  <=> (-5)³⁰ < (-3)⁵⁰

99²⁰=(99²)¹⁰=9801¹⁰  do 9801<9999 <=> 9801¹⁰ < 9999¹⁰   <=>  99²⁰ < 9999¹⁰

21¹²=(21³)⁴=9261⁴  9261>54  => 9261⁴ > 54⁴   <=>  21¹² > 54⁴

4444³³³³=((4*1111)³³³³=4³³³³ * 1111³³³³=64¹¹¹¹ * 1111³³³³          3333⁴⁴⁴⁴=3⁴⁴⁴⁴ * 1111⁴⁴⁴⁴=81¹¹¹¹ * 1111⁴⁴⁴⁴ do 64¹¹¹¹ < 81¹¹¹¹ va 1111³ < 1111⁴ nen 4444³³³³ < 3333⁴⁴⁴⁴

a, Ta có: 5^30 = (5^3)^10= 125^ 10    >   (-10^2)^10= 100^10

b, ta có: 21^12= ( 21^3)^4 > 54^4

 c, Ta có: (1/16)^10 = 1/16^10

                 (1/2)^50= 1/2^50

   Lại có: 16^10=(2^4)^10= 2^40 < 2^50 nên (1/6)^10> (1/2)^50

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\(a:ta.c\text{ó}:BCNN:12\\ \dfrac{2}{3}=\dfrac{2\cdot4}{3\cdot4}=\dfrac{8}{12};\dfrac{1}{4}=\dfrac{1\cdot3}{4\cdot3}=\dfrac{3}{12}\\ v\text{ì }\dfrac{8}{12}< \dfrac{3}{12}n\text{ê}n\dfrac{2}{3}< \dfrac{1}{4}\\ b:ta.c\text{ó}:\\ 10=2\cdot5\\ 8=2^3\\ \Rightarrow BCNN=2^3\cdot5=8\cdot5=40\\ \dfrac{7}{10}=\dfrac{7\cdot4}{10\cdot4}=\dfrac{28}{40};\dfrac{7}{8}=\dfrac{7\cdot5}{8\cdot5}=\dfrac{35}{40}\\ v\text{ì }\dfrac{28}{40}< \dfrac{35}{40}n\text{ê}n\dfrac{7}{10}< \dfrac{7}{8}\\ c:ta.c\text{ó}:\\ 7=7;5=5\\ \Rightarrow BCNN=7\cdot5=35\\ \dfrac{6}{7}=\dfrac{6\cdot5}{7\cdot5}=\dfrac{30}{35};\dfrac{3}{5}=\dfrac{3\cdot7}{5\cdot7}=\dfrac{21}{35}\\ v\text{ì }\dfrac{30}{35}>\dfrac{21}{35}n\text{ê}n\dfrac{6}{7}>\dfrac{3}{5}\\ d:ta.c\text{ó}:\\ 21=3\cdot7\\ 72=2^3\cdot3^2\\ \Rightarrow BCNN=2^3\cdot3^2\cdot7=504\\ \dfrac{14}{21}=\dfrac{14\cdot24}{21\cdot24}=\dfrac{336}{504};\dfrac{60}{72}=\dfrac{60\cdot7}{72\cdot7}=\dfrac{420}{504}\\ v\text{ì }\dfrac{336}{504}< \dfrac{420}{504}n\text{ê}n\dfrac{14}{21}< \dfrac{60}{72}\)

\(a,\Rightarrow2A=2+2^2+...+2^{2011}\)

\(\Rightarrow2A-A=2+2^2+...+2^{2011}-2^0-2-..-2^{2010}\)

\(\Rightarrow A=2^{2011}-1=B\)

\(b,A=2019.2011=\left(2010-1\right)\left(2010+1\right)=\left(2010-1\right).2010+\left(2010-1\right)=2010^2-2010+2010-1=2010^2-1< 2010^2=B\)

\(a,\Rightarrow2A=2^1+2^2+...+2^{2011}\\ \Rightarrow2A-A=A=2^{2011}-2^0=2^{2011}-1=B\)

\(b,A=\left(2010-1\right)\left(2010+1\right)=2010^2+2010-2010-1=2010^2-1< 2010^2=B\)

a) \(\frac{{ - 21}}{{10}}\) < 0

b) \(\frac{{ - 5}}{{ - 2}} = \frac{5}{2} > 0\). Vậy \(\frac{{ - 5}}{{ - 2}} > 0\).

c) \(\frac{{ - 5}}{{ - 2}} = \frac{5}{2} > 0\), mà \(\frac{{ - 21}}{{10}} < 0\)

Vậy \(\frac{{ - 5}}{{ - 2}} > \frac{{ - 21}}{{10}}\).

a: \(-\dfrac{21}{10}< 0\)

b: \(0< -\dfrac{5}{-2}\)

c: \(-\dfrac{21}{10}< 0< \dfrac{-5}{-2}\)

a)Ta có 2222^3333=2222^3x1111=(2222^3)^1111=(1111^3x2^3)^1111=(1111^3x8)^1111

Tương tự:ta có:3333^2222=(1111^3x9)^1111

Vì 8<9 nên 2222^3333<3333^2222

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b>

c>

d>