gọi ps đầu là A

gọi ps tiếp đó là B

Tôi khẳng định là A<B

a)  23 21 > 1 > 21 23 ⇒ 23 21 > 21 23

b)  311 256 > 1 > 199 203 ⇒ 311 256 > 199 203 .

a ) Ta có :

\(1-\frac{41}{91}=\frac{50}{91}\) \(=\frac{500}{910}\)  ;  \(1-\frac{411}{911}=\frac{500}{911}\)

 Vì \(\frac{500}{910}>\frac{500}{911}\)nên \(\frac{41}{91}< \frac{411}{911}\)

b ) Ta có :

\(1-\frac{113}{115}=\frac{2}{115}\)   ;     \(1-\frac{93}{95}=\frac{2}{95}\)

Vì \(\frac{2}{115}< \frac{2}{95}\)nên \(\frac{113}{115}>\frac{93}{95}\).

c ) Quy đồng TS ta có :

\(\frac{13}{53}=\frac{143}{583}\) ;   \(\frac{11}{30}=\frac{143}{390}\)

Vì \(\frac{143}{583}< \frac{143}{390}\)nên   \(\frac{13}{53}< \frac{11}{30}\).

a: \(2^{\dfrac{6}{3}}=2^2\)

b: \(2^{\dfrac{6}{3}}=2^2=4\)

\(\sqrt[3]{2^6}=\sqrt[3]{64}=4\)

=>\(2^{\dfrac{6}{3}}=\sqrt[3]{2^6}\)

a) $\frac{2}{3} = \frac{{2 \times 6}}{{3 \times 6}} = \frac{{12}}{{18}}$

Ta có $\frac{{12}}{{18}} > \frac{{11}}{{18}}$ nên $\frac{2}{3} > \frac{{11}}{{18}}$

b) $\frac{{36}}{{63}} = \frac{{36:9}}{{63:9}} = \frac{4}{7}$

Ta có $\frac{4}{7} < \frac{5}{7}$ nên $\frac{{36}}{{63}}$ < $\frac{5}{7}$

c)

$\frac{{55}}{{110}} = \frac{{55:55}}{{110:55}} = \frac{1}{2}$ ;  $\frac{4}{8} = \frac{1}{2}$

Vậy $\frac{{55}}{{110}}$ = $\frac{4}{8}$

Ta có:

\(\frac{{ - 7}}{{18}} = \frac{{ - 7.2}}{{18.2}} = \frac{{ - 14}}{{36}}\)

\(\frac{5}{{ - 12}} = \frac{{ - 5}}{{12}} = \frac{{ - 5.3}}{{12.3}} = \frac{{ - 15}}{{36}}\)

Vì \(\frac{{ - 14}}{{36}} > \frac{{ - 15}}{{36}}\) nên \(\frac{{ - 7}}{{18}} > \frac{5}{{ - 12}}\).

Cách 1:

Ta có: \(\frac{{ - 4}}{{ - 5}} = \frac{4}{5}\)  và \(\frac{2}{{ - 5}} = \frac{{ - 2}}{5}\)

Do \(4 >  - 2\) nên \(\frac{4}{5} > \frac{{ - 2}}{5}\)

Cách 2:

Ta có: \(\frac{{ - 4}}{{ - 5}} = \frac{4}{5} > 0\) và \(\frac{2}{{ - 5}} < 0\)

\( \Rightarrow \frac{{ - 4}}{{ - 5}} > \frac{2}{{ - 5}}\).

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

a)      Ta có \(\frac{{ - 2}}{3} < 0\) và \(\frac{1}{{200}} > 0\) nên \(\frac{{ - 2}}{3}\)<\(\frac{1}{{200}}\).

b)      Ta có: \(\frac{{139}}{{138}} > 1\) và \(\frac{{1375}}{{1376}} < 1\) nên \(\frac{{139}}{{138}}\) > \(\frac{{1375}}{{1376}}\).

c)      Ta có: \(\frac{{ - 11}}{{33}} = \frac{{ - 1}}{3}\) và \(\frac{{25}}{{ - 76}} = \frac{{ - 25}}{{76}} > \frac{{ - 25}}{{75}} = \frac{{ - 1}}{3}\,\,\,\, \Rightarrow \frac{{25}}{{ - 76}} > \frac{{ - 11}}{33}\).

a: -2/3<0<1/200

b: 139/138>1

1375/1376<1

=>139/138>1375/1376

c: -11/33=-1/3=-25/75<-25/76