Lời giải:

a.

$32^{47}=(2^5)^{47}=2^{5.47}=2^{235}$

$64^{33}=(2^6)^{33}=2^{6.33}=2^{198}$

Vì $2^{235}> 2^{198}$ nên $32^{47}> 64^{33}$

b.

$(\frac{1}{2})^{30}=\frac{1}{2^{30}}=\frac{1}{8^{10}}$

$(\frac{1}{3})^{20}=\frac{1}{3^{20}}=\frac{1}{9^{10}}$

Hiển nhiên $8^{10}< 9^{10}\Rightarrow \frac{1}{8^{10}}> \frac{1}{9^{10}}$

$\Rightarrow (\frac{1}{2})^{30}> (\frac{1}{3})^{20}$

`3/7-2/5`

`=1/35>0`

`=>3/7>2/5`

`b,9>8`

`=>1/9<1/8`

`=>5/9<5/8`

`d,8/7>1`

`7/8<1`

`=>8/7>7/8`

a)5/8 < 8/9

b)8/12> 5/9

c)7/12 < 11/18

a)5/8 < 8/9

b)8/12 > 5/9

c)7/12 < 11/18

a) Vì 39 < 54 nên -39 > -54

b) Vì 3 179 < 3 279 nên - 3 179 > - 3 279.

`a)1<3`

`=>1/5<3/5`

`b)21>9`

`=>8/21<8/9`

`c)3/5<5/5=1`

`d)7/5>5/5=1`

a)3/5>1/5 b)8/21<8/9 c)3/5<1 d)7/5>1

a: \(\dfrac{-8}{31}=\dfrac{-8\cdot101}{31\cdot101}=\dfrac{-808}{3131}\)

\(\dfrac{-789}{3131}=\dfrac{-789}{3131}\)

b: Thiếu phân số thứ hai rồi bạn

c: \(\dfrac{1}{n}=\dfrac{n+1}{n\left(n+1\right)}\)

\(\dfrac{1}{n+1}=\dfrac{n}{n\left(n+1\right)}\)

a: so sánh với 1

64/85 < 73/81

b: so sánh với 1 

n + 1/n+2 > n/ n+3

c: so sánh với 1

64/65 > 60/61

d: so sánh với 1

99/97 < 88/86

Câu b thì gg search nhé

\(3.\dfrac{4}{10}< 3.\dfrac{9}{10}\\ 5.\dfrac{1}{10}>2.\dfrac{9}{10}\\ 3.\dfrac{4}{10}=3.\dfrac{2}{5}\)

a) \(3\cdot\dfrac{4}{10}< 3\cdot\dfrac{9}{10}\)

b) \(5\cdot\dfrac{1}{10}< 2\cdot\dfrac{9}{10}\)

c) \(3\cdot\dfrac{4}{10}=3\cdot\dfrac{2}{5}\)

 a) \(\frac{{ - 3}}{8} = \frac{{ - 3.3}}{{8.3}} = \frac{{ - 9}}{{24}}\)

Vì -9 < -5 nên \(\frac{{ - 9}}{{24}} < \frac{{ - 5}}{{24}}\)

Vậy \(\frac{{ - 3}}{8} < \frac{{ - 5}}{{24}}\).

b) Cách 1: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5}; \frac{3}{{ - 5}} = \frac{-3}{{5}}\)

Vì 2 > -3 nên \(\frac{2}{5} > \frac{-3}{{5}}\)

Vậy \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).

Cách 2: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5} > 0\) mà \(\frac{3}{{ - 5}} < 0\)

\(\Rightarrow\) \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).

c) \(\frac{{ - 3}}{{ - 10}} = \frac{3}{{10}} = \frac{{3.2}}{{10.2}} = \frac{6}{{20}}\)

\(\frac{{ - 7}}{{ - 20}} = \frac{7}{{20}}\)

Vì 6 < 7 nên \(\frac{6}{{20}} < \frac{7}{{20}}\) nên \(\frac{{ - 3}}{{ - 10}} < \frac{{ - 7}}{{ - 20}}\).

d) \(\frac{{ - 5}}{4} = \frac{{ - 5.5}}{{4.5}} = \frac{{ - 25}}{{20}}; \frac{{ 23}}{{-20}}=\frac{{-23}}{{20}} \)

Vì -25 < -23 nên \( \frac{{ - 25}}{{20}} < \frac{{-23}}{{20}} \)

Vậy \(\frac{{ - 5}}{4} < \frac{{23}}{{ - 20}}\).