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290=(25)18=3218
536=(52)18=2518
Vì 32>25 nên 3218>2518
=>290>536
b,15=\(\sqrt{225}\) <\(\sqrt{235}\)
=> 15<\(\sqrt{235}\)
c, Ta có: \(\dfrac{1}{3}=\dfrac{13}{39}\)
vì 38<39
nên \(\dfrac{13}{38}>\dfrac{13}{39}\)
a) 290= (210)9mà 210=(25)2
536= (54)9mà 54=(52)2
Do 25>52nên 290>536
a) \(\dfrac{13}{38}\) và \(\dfrac{1}{3}\)
\(\dfrac{1}{3}\) = \(\dfrac{13}{39}\) < \(\dfrac{13}{38}\)
=> \(\dfrac{13}{38}>\dfrac{1}{3}\)
b)\(\sqrt{235}\) và 15
15 = \(\sqrt{225}\) < \(\sqrt{235}\) ( vì 225 < 235)
=> \(\sqrt{235}>15\)
tick mình nha
=>
a, Ta có:
\(\dfrac{13}{38}\)=\(\dfrac{39}{114}\) ; \(\dfrac{1}{3}\)=\(\dfrac{38}{114}\)
Vì 38 < 39 ⇒ \(\dfrac{39}{114}>\dfrac{38}{114}\)
Vay \(\dfrac{13}{38}>\dfrac{1}{3}\)
b, Goi \(\sqrt{235}\)= a ⇒ 235 = \(a^2\)
Ta có : 15^2= 225
Vì 235 > 225 nên a^2 > 15^2
Vay \(\sqrt{235}\)>15
a) Ta có : \(\dfrac{-1}{5}< 0< \dfrac{1}{1000}\)
\(\Rightarrow\dfrac{-1}{5}< \dfrac{1}{1000}\)
b) Ta có : \(\dfrac{267}{268}< 1< \dfrac{1347}{1343}\)
=> \(\dfrac{267}{-268}< -\dfrac{1347}{1343}\)
c) \(\dfrac{13}{38}>\dfrac{13}{39}=\dfrac{1}{3}=\dfrac{19}{87}>\dfrac{29}{88}\)
=> \(-\dfrac{13}{38}< \dfrac{29}{-88}\)
d) \(\dfrac{181818}{313131}=\dfrac{18}{31}\)
=> \(-\dfrac{18}{31}=-\dfrac{181818}{313131}\)
So sánh các số hữu tỉ:
a) và
b) và
c) x = -0,75 và
Lời giải:
a)
Vì -22 < -21 và 77> 0 nên x <y
b)
Vì -216 < -213 và 300 > 0 nên y < x
c)
Vậy x=y
Lời giải:
a)
Vì -22 < -21 và 77> 0 nên x <y
b)
Vì -216 < -213 và 300 > 0 nên y < x
c)
Vậy x=y
a) \(3^{21}\)và \(2^{31}\)
\(3^{21}\)=\(3.3^{20}\)=\(3.9^{10}\)
\(2^{31}=2.2^{30}=2.8^{10}\)
Vì \(3.9^{10}\)>\(2.8^{10}\)\(\Rightarrow3^{21}>2^{31}\)
b)\(2^{300}\)và \(3^{200}\)
\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
Vì \(8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)
c)\(32^9\)và\(18^{13}\)
\(32^9=2^{5.9}=2^{45}\)
\(18^{13}>16^{13}=2^{4.13}=2^{52}\)
\(\Rightarrow2^{45}< 2^{52}< 18^{13}\)\(\Rightarrow2^{45}< 18^{13}\Rightarrow32^9< 18^{13}\)
a) ta có: 321 = 3.320 = 3.910
231 = 2.230 = 2.810
vì 2.810 < 3.910 => 231 < 321
b) ta có: 2300 = (23)100 = 8100
3200 = (32)100 = 9100
vì 8100 < 9100 => 2300 < 3200
c) ta có: 329 = (25)9 = 245
1813 > 1613 = (24)13 = 252
ta thấy 245 < 252 < 1813
Nên 329 < 1813
B1
a. = 7/3. ( 37/5 - 32/5)
= 7/3 . 1
= 7/3
Phần b có gì đó sai sao lại có 3:+
c. = 4 + 6 - 3 + 5
= 12
d. = -5/21 : -19/21 : 4/5
= 25/76
B2
a. 1/4 : x =1/2 - 3/4
x = -1/4
x = 1/4 : -1/4
x = -1
b. 2 . | 2x - 3 | = 4 - (-8)
2 . | 2x - 3| = 12
| 2x - 3 | = 12:2
| 2x - 3 | = 6
| x - 3 | = 6:2
| x - 3 | = 3
=> x - 3 = +- 3
* x - 3 = 3
x = 6
* x - 3 = -3
x = 0
Chúc bạn vui vẻ 
\(a)\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{12}{24}+\left(-\dfrac{5}{24}\right)\)
\(=\dfrac{18+12+\left(-5\right)}{24}\)
\(=\dfrac{25}{24}\)
\(b)\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)
\(=\dfrac{5}{7}.\dfrac{-2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)
\(=\dfrac{5}{7}\left(\dfrac{-2}{13}+\dfrac{-11}{13}+\dfrac{13}{13}\right)\)
\(=\dfrac{5}{7}.0=0\)
\(c)\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)
\(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}\)
\(=2\dfrac{1}{2}\)
\(d)\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}\)
\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{38}{51}+\dfrac{306}{714}\)
\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{532}{714}+\dfrac{306}{714}\)
\(=\dfrac{1391}{714}\)
a)\(\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}=\dfrac{18}{24}+\dfrac{12}{24}-\dfrac{5}{24}=\dfrac{25}{24}\)
b)\(\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}=\dfrac{5}{7}\left(\dfrac{-2}{13}-\dfrac{11}{13}+1\right)=\dfrac{5}{7}.0=0\)
c)\(\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}=1+1+\dfrac{1}{2}=2,5\)
d)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}=\dfrac{15}{34}+\left(\dfrac{1}{3}+\dfrac{38}{51}+\dfrac{3}{7}\right)=\dfrac{15}{34}+\dfrac{538}{357}=\dfrac{1391}{714}\)
a) 45<1<1,1⇒45<1,145<1<1,1⇒45<1,1
b) -500 < 0 < 0,001 => -500 < 0,001
c) −12−37=1237<1236=13=1339<1338⇒−12−37<1338
a)Ta có :
\(\dfrac{4}{5}< 1< 1,1\Rightarrow\dfrac{4}{5}< 1,1\)
b)Ta có :
\(-500< 0< 0,001\Rightarrow-500< 0,001\)
c)Ta có :
\(\dfrac{-12}{-37}=\dfrac{12}{37}< \dfrac{12}{36}=\dfrac{1}{3}=\dfrac{13}{39}< \dfrac{13}{38}\Rightarrow\dfrac{-12}{-37}< \dfrac{13}{38}\)
a) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
Do : \(8^{100}< 9^{100}\)
=> \(2^{300}< 3^{200}\)
b) Do \(\dfrac{13}{38}>\dfrac{13}{39}\)
Mà : \(\dfrac{13}{39}=\dfrac{1}{3}\)
=> \(\dfrac{13}{38}>\dfrac{1}{3}\)
c)Do : \(\sqrt{235}>\sqrt{225}\)
Mà : \(\sqrt{225}=15\)
=> \(\sqrt{235}>15\)
a) Ta có:
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
Ta thấy 8<9 suy ra \(8^{100}< 9^{100}\)
Vậy \(2^{300}< 3^{200}\)