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Bài 1:
a) Ta có: \(13A=\dfrac{13^{16}+13}{13^{16}+1}=1+\dfrac{12}{13^{16}+1}\)
\(13B=\dfrac{13^{17}+13}{13^{17}+1}=1+\dfrac{12}{13^{17}+1}\)
Vì \(\dfrac{12}{13^{16}+1}>\dfrac{12}{13^{17}+1}\Rightarrow1+\dfrac{12}{13^{16}+1}>1+\dfrac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\)
\(\Rightarrow A>B\)
Vậy A > B
b) Ta có: \(1999C=\dfrac{1999^{2000}+1999}{1999^{2000}+1}=1+\dfrac{1998}{1999^{2000}+1}\)
\(1999D=\dfrac{1999^{1999}+1999}{1999^{1999}+1}=1+\dfrac{1998}{1999^{1999}+1}\)
\(\dfrac{1998}{1999^{2000}+1}< \dfrac{1998}{1999^{1999}+1}\Rightarrow1+\dfrac{1998}{1999^{2000}+1}< 1+\dfrac{1999}{1999^{1999}+1}\)
\(\Rightarrow1999C< 1999D\)
\(\Rightarrow C< D\)
Vậy C < D
Ta có 22/37 < 29/37 và 29/37 < 29/33
=> 22/37< 29/37 < 29/33
a, 15.27 - 3.5.17
= 15.27 - 15. 17
= 15 . ( 27 - 17 )
= 15 . 10
= 150
b, 55 - 5 . ( 20 + 11 )
= 55 - 100 . 55
= 55 . ( 1 - 100 )
= 55 . ( -99 )
= ...
Ta có :\(37.\left(29-23\right)-29.\left(37-23\right)=37.29-37.23-29.37+29.23\)
\(=23.29-37.23\)
\(=23.\left(29-37\right)\)
\(=23.\left(-8\right)\)
\(=-184\)
37. (29 − 23) − 29.(37 − 23)
=[ 37.29 − 37.23 ] − [ 29.37 − 29.23 ]
=(37.29)-(37.23)-(29.37)+(29.23)
=(29.23)-(37.23)
=23(29-37)
=23.(-8)
=-184
37 . (29 - 23) - 29. (37 - 23)
= 37 . 29 - 37 . 23 - 29 . 37 - 29 . 23
= 37 . (29 - 29) - 23 . (37 - 29)
= 37 . 0 - 23 . 8
= 0 - 184
= 0 + (-184)
= -184
Hok Tốt
a ) Ta có
\(\frac{29}{33}>\frac{29}{37}\)( đồng tử khác mẫu )
\(\frac{22}{37}< \frac{29}{37}\)( đồng mẫu khác tử )
=> \(\frac{29}{33}>\frac{29}{37}>\frac{22}{37}\)
b ) \(\frac{163}{257}< \frac{163}{221}\)
\(\frac{162}{257}>\frac{149}{257}\)
\(\Rightarrow\frac{163}{221}>\frac{163}{257}>\frac{149}{257}\)
a) ta có: \(\frac{22}{37}< \frac{29}{37}\)
\(\frac{29}{33}>\frac{29}{37}\)
\(\Rightarrow\frac{22}{37}< \frac{29}{37}< \frac{29}{33}\)
b) ta có: \(\frac{163}{257}>\frac{149}{257}\)
\(\frac{163}{221}>\frac{163}{257}\)
\(\Rightarrow\frac{163}{221}>\frac{163}{257}>\frac{149}{257}\)