ai giúp mink với ;so sánh hai phân số -3/20 và -6/20
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Ta có : 6/10 = 6.2/10.2 = 12/20
Vì 11/20 < 12/20 ( 11 < 12 ) nên 11/20 < 6/10
Vậy 11/20 < 6/10

Ta có: \(\dfrac{6}{{15}} = \dfrac{{6:3}}{{15:3}} = \dfrac{2}{5}\)
\(\begin{array}{l}BCNN\left( {20,5} \right) = 20\\\dfrac{2}{5} = \dfrac{{2.4}}{{5.4}} = \dfrac{8}{{20}}\end{array}\)
Vì 3 < 8 nên \(\dfrac{3}{{20}} < \dfrac{8}{{20}}\)
Suy ra \(\dfrac{3}{{20}} < \dfrac{6}{{15}}\)

a. 18/72 = 1/4
15/20 = 3/4
vì 1/4<3/4 nên 18/72<15/20
b. 24/36 = 2/3
20/50 = 2/5
2/3>2/5 nên 24/36>20/50
a, \(\dfrac{18}{72}\) = \(\dfrac{18:18}{72:18}\) = \(\dfrac{1}{4}\); \(\dfrac{15}{20}\) = \(\dfrac{3}{4}\)
Vì \(\dfrac{1}{4}\) < \(\dfrac{3}{4}\) nên \(\dfrac{18}{72}\) < \(\dfrac{15}{20}\)
b, \(\dfrac{24}{36}\) = \(\dfrac{24:12}{36:12}\) = \(\dfrac{2}{3}\); \(\dfrac{20}{50}\) = \(\dfrac{20:10}{50:10}\) = \(\dfrac{2}{5}\)
Vì \(\dfrac{2}{3}\) > \(\dfrac{2}{5}\) nên \(\dfrac{24}{36}\) > \(\dfrac{20}{50}\)


Ta có: \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\Rightarrow A< B\)
Vậy A < B

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

dễ mà :
1, 4/6=2/3
18/27=2/3
nen 4/6=18/27
2,12/20=3/5
22/55=2/5
vi 3/5 > 2/5
nen 12/20>22/55
3, 5/6=1-1/6
1111/1212=1-101/1212
vi 1/6>101/1212 nen 5/6<1111/1212

a) (x - 3)(y - 3) = 9 = 1.9 = 3.3
Lập bảng:
x - 3 | 1 | -1 | 3 | -3 | 9 | -9 |
y - 3 | 9 | -9 | 3 | -3 | 1 | -1 |
x | 4 | 2 | 6 | 0 | 12 | -3 |
y | 12 | -6 | 6 | 0 | 4 | 2 |
Vậy ...
b) A = \(\frac{10^{19}+1}{10^{20}+1}\) => 10A = \(\frac{10^{20}+10}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)
B = \(\frac{10^{20}+1}{10^{21}+1}\) => 10B = \(\frac{10^{21}+10}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)
Do \(10^{20}+1< 10^{21}+1\) => \(\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\) => 10A > 10B => A > B

a/3^34=(3^3)^11 x 3
=27^11 x 3
5^20 = (5^2)^10
= 25^10
có 27^11 x3> 25^10(27>25 và 11>10)
suy ra 3^34>5^20
b/17^20=(17^2)^10
=289^10
có 289>71 ; 10>5
nên 71^5>17^20
Toán lớp 6 mà
\(-\frac{3}{20}và-\frac{6}{20}\)
ta có \(\frac{3}{20}<\frac{6}{20}\) =>\(\frac{-3}{20}>-\frac{6}{20}\)