a) \(\dfrac{-1}{20}=\dfrac{-7}{140}\)

\(\dfrac{5}{7}=\dfrac{100}{140}\)

mà -7<100

nên \(-\dfrac{1}{20}< \dfrac{5}{7}\)

b) \(\dfrac{216}{217}< 1\)

\(1< \dfrac{1164}{1163}\)

nên \(\dfrac{216}{217}< \dfrac{1164}{1163}\)

c) \(\dfrac{-12}{17}=\dfrac{-180}{255}\)

\(\dfrac{-14}{15}=\dfrac{-238}{255}\)

mà -180>-238

nên \(-\dfrac{12}{17}>\dfrac{-14}{15}\)

d) \(\dfrac{27}{29}>0\)

\(0>-\dfrac{2727}{2929}\)

nên \(\dfrac{27}{29}>-\dfrac{2727}{2929}\)

a: \(\sqrt{12}<\sqrt{12,25}=3,5\)

\(\sqrt{20}<\sqrt{20,25}=4,5\)

\(\sqrt{30}<\sqrt{30,25}=5,5\)

\(\sqrt{42}<\sqrt{42,25}=6,5\)

Do đó: \(\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\) <3,5+4,5+5,5+6,5

=>A<20

b: \(B=\sqrt{196}-\frac{1}{\sqrt3}-1=14-\frac{1}{\sqrt3}-1=13-\frac{1}{\sqrt3}\)

\(C=\sqrt{169}+\frac{1}{-\sqrt2}=13-\frac{1}{\sqrt2}\)

3>2

=>\(\frac{1}{\sqrt3}<\frac{1}{\sqrt2}\)

=>\(-\frac{1}{\sqrt3}>-\frac{1}{\sqrt2}\)

=>\(-\frac{1}{\sqrt3}+13>-\frac{1}{\sqrt2}+13\)

=>B>C

a. 7 . 10 > 0

b. 123 . 8 > 12 . 31

c. 15 . 28 < 22 . 27

d. 17 . 3 > 23 . 2 

HT nha^^

7.0  >  0                      123.8  > 12.31                     15.28  <  22.27                         17.  > 23.2

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

b: 3^500=243^100

5^300=125^300

=>3^500>5^300

\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)

\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)

Bài 2: 

a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)

\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)

\(=-\dfrac{3}{5}\)

b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)

\(\Leftrightarrow8x-1=5\)

\(\Leftrightarrow8x=6\)

hay \(x=\dfrac{3}{4}\)

\(\dfrac{13}{17}< \dfrac{30}{34}\)                                        \(\dfrac{9}{8}>\dfrac{5}{7}\) 

 \(\dfrac{15}{7}< \dfrac{8}{3}\)                                        \(\dfrac{5}{7}< \dfrac{7}{5}\)                     

\(\dfrac{16}{20}=\dfrac{4}{5}\)

b: \(\dfrac{3}{\sqrt{7}-2}-\dfrac{4}{\sqrt{7}+\sqrt{3}}\)

\(=\sqrt{7}+2-\sqrt{7}+\sqrt{3}=2+\sqrt{3}\)

   \(\dfrac{3}{17}\) .\(\dfrac{6}{29}\) - \(\dfrac{3}{17}\).\(\dfrac{35}{29}\) + 2022\(\dfrac{3}{17}\)

= (\(\dfrac{3}{17}\).\(\dfrac{6}{29}\) - \(\dfrac{3}{17}\).\(\dfrac{35}{29}\)) + 2022 + \(\dfrac{3}{17}\)

= \(\dfrac{3}{17}\).(\(\dfrac{6}{29}\) - \(\dfrac{35}{29}\)) + 2022 + \(\dfrac{3}{17}\)

= \(\dfrac{3}{17}\).(-1) + 2022 + \(\dfrac{3}{17}\)

 = (- \(\dfrac{3}{17}\) + \(\dfrac{3}{17}\)) + 2022

= 0 + 2022

= 2022 

b; \(\dfrac{5}{6}\) - (\(\dfrac{1}{3}\) + \(\dfrac{1}{2}\)).20%

= \(\dfrac{5}{6}\) - \(\dfrac{5}{6}\).\(\dfrac{1}{5}\)

= \(\dfrac{5}{6}-\dfrac{1}{6}\)

= \(\dfrac{4}{6}\) 

= \(\dfrac{2}{3}\)