So sánh
1/ 3 và √15 -1
2/ 6 và √26 +1
3/ √2 -2 và √3 -3
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1) `-3\sqrt13=-3\sqrt13`
`-9=-3\sqrt9`
`\sqrt13>\sqrt9`
`=> -3\sqrt13 < -3\sqrt9`
`=> -3\sqrt13 < 9`.
2) `\sqrt15 < \sqrt16`
`<=> \sqrt15-1 < \sqrt16-1`
`<=> \sqrt15-1 < 3 < \sqrt10`
`=> \sqrt15-1 <\sqrt10`
3) `5=4+1=\sqrt16+1`
`\sqrt8+1=\sqrt8+1`
`=> 5>\sqrt8+1`
1) \(-3\sqrt{13}=-\sqrt{117}< -\sqrt{81}=-9\)
3) Ta có: \(5^2=25=9+16\)
\(\left(2\sqrt{2}+1\right)^2=9+4\sqrt{2}\)
mà \(16>4\sqrt{2}\)
nên \(5>2\sqrt{2}+1\)
Giải:
a) \(11\dfrac{3}{4}.\left(6\dfrac{5}{6}-4\dfrac{1}{2}+1\dfrac{2}{3}\right)\)
\(=\dfrac{47}{4}.\left(\dfrac{41}{6}-\dfrac{9}{2}+\dfrac{5}{3}\right)\)
\(=\dfrac{47}{4}.4\)
\(=47\)
b) \(\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\)
\(=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)
\(=\dfrac{25}{8}:\dfrac{75}{26}\)
\(=\dfrac{13}{12}\)
c) \(\left(17\dfrac{13}{15}-3\dfrac{3}{7}\right)-\left(2\dfrac{12}{15}-4\right)\)
\(=\dfrac{268}{15}-\dfrac{24}{7}-\dfrac{14}{5}+4\)
\(=\left(\dfrac{268}{15}-\dfrac{14}{5}\right)+\left(\dfrac{-24}{7}+4\right)\)
\(=\dfrac{226}{15}+\dfrac{4}{7}\)
\(=\dfrac{1642}{105}\)
d) \(2\dfrac{2}{3}.\left(\dfrac{-4}{5}.0,375.-10.\dfrac{-15}{24}\right)\)
\(=\dfrac{8}{3}.\left(\dfrac{-4}{5}.\dfrac{3}{8}.-10.\dfrac{-5}{8}\right)\)
\(=\left(\dfrac{8}{3}.\dfrac{3}{8}\right).\left(\dfrac{-4}{5}.\dfrac{-5}{8}.-10\right)\)
\(=1.-5\)
\(=-5\)
Chúc bạn học tốt!
a) 12/17 và 7/153
=>12/17 = 108/153
=>108/153 > 7/153
Vậy 12/17 > 7/153
b) Vì : 1999/2001 < 1 và 12/11 > 1 nên 1999/2001 < 12/11
c) 13/60 và 27/100
13/60 < 15/60 = 1/4
27/100 > 25/100 = 1/4
vậy 13/60 < 27/100
d) Ta có: 1 - 13/27 = 14/27
1 - 27/41 = 14/41
Vì 14/27 > 14/41 nên 13/27 < 27/41
bài làm
a) 12/17 và 7/153
=>12/17 = 108/153
=>108/153 > 7/153
Vậy 12/17 > 7/153
b) Vì : 1999/2001 < 1 và 12/11 > 1 nên 1999/2001 < 12/11
c) 13/60 và 27/100
13/60 < 15/60 = 1/4
27/100 > 25/100 = 1/4
vậy 13/60 < 27/100
d) Ta có: 1 - 13/27 = 14/27
1 - 27/41 = 14/41
Vì 14/27 > 14/41 nên 13/27 < 27/41
Bài 1:
\(\dfrac{-5}{18}=\dfrac{-20}{72};\dfrac{7}{-24}=\dfrac{-21}{72}.\)
\(\dfrac{-15}{-40}=\dfrac{3}{8}=\dfrac{9}{24};\dfrac{24}{-72}=\dfrac{-1}{3}=\dfrac{-8}{24}.\)
Bài 3:
a) \(\dfrac{2}{3}h=\dfrac{8}{12}h;\dfrac{3}{4}h=\dfrac{9}{12}h.\Rightarrow\dfrac{2}{3}h< \dfrac{3}{4}h.\)
b) \(\dfrac{4}{5}km/h=\dfrac{8}{10}km/h;\dfrac{9}{10}km/h.\Rightarrow\dfrac{4}{5}km/h< \dfrac{9}{10}km/h.\)
\(\dfrac{3}{17}\) .\(\dfrac{6}{29}\) - \(\dfrac{3}{17}\).\(\dfrac{35}{29}\) + 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%
= \(\dfrac{5}{6}\) - \(\dfrac{5}{6}\).\(\dfrac{1}{5}\)
= \(\dfrac{5}{6}-\dfrac{1}{6}\)
= \(\dfrac{4}{6}\)
= \(\dfrac{2}{3}\)
`1)1/2:2/3 .... 2/3 : 1/2`
`=>1/2xx3/2 .... 2/3xx2`
`=>3/4 .... 4/3`
Vì `3/4 < 1` và `4/3>1`
`=>3/4<4/3`
__
`4/7:2/5 ... 4/7 : 3/5`
`=>4/7xx5/2....4/7xx5/3`
`=>20/14...20/21`
`=>10/7...20/21`
Vì `10/7>1` và `20/21<1`
`=>10/7>20/21`
__
`4/15:4/7....2/5xx10/3`
`=>4/15xx7/4...20/15`
`=>7/15...20/15`
Vì `7<20` nên `7/15<20/15`
__
`5/6...15/18-11/18`
`=>5/6...4/18`
Ta có : MSC : `18`
`5/6 = 15/18`
Vì `15>4` nên `5/6 > 4/18`
a) 13^1 <13^15
b)7^3 < 8^3
c) 7 - 6 = 13-12
d) 2^300 <3^200
1 \(\sqrt{15}< \sqrt{16}=4\Rightarrow\sqrt{15}-1< 4-1=3\Rightarrow3>\sqrt{15}-1\)
2 \(\sqrt{26}>\sqrt{25}=5\Rightarrow\sqrt{26}+1>5+1=6\Rightarrow6< \sqrt{26}+1\)
3 \(\sqrt{2}>\sqrt{1}=1\Rightarrow\sqrt{2}-2>1-2=-1\left(1\right)\)
\(\sqrt{3}< \sqrt{4}=2\Rightarrow\sqrt{3}-3< 2-3=-1\left(2\right)\)
từ \(\left(1\right)\left(2\right)\Rightarrow\sqrt{2}-2>\sqrt{3}-3\)