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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
a) 820 và 720
vì 8>7 nên 820>720
b) 420 và 1620
vì 4<16 nên 420<1620
c) 277= (33)7= 321
815=( 34)5=320
vì 21>20 nên 321>320 hay 277> 815
e) 521= 520 . 5
vì 520 . 5>520 . 4 nên 521> 4 . 520
Bài 1 :
a,820 > 720
b, 420 = 1610
c, 277 > 815
d , 554 > 381
e, 521 > 4 . 520
f, 220 > 7.217
a) Giải
So sánh từng số hạng của A với B, ta thấy:
\(\dfrac{19}{41}< \dfrac{21}{41};\dfrac{23}{53}< \dfrac{23}{49}\) và \(\dfrac{29}{61}< \dfrac{33}{65}\) (vì 29.65 < 33.61)
\(\Rightarrow\dfrac{19}{41}+\dfrac{23}{53}+\dfrac{29}{61}< \dfrac{21}{41}+\dfrac{23}{49}+\dfrac{33}{65}\)
\(\Rightarrow A< B\)
Vậy A < B
b) Giải
Ta có: \(C=\dfrac{19^{20}+5}{19^{20}-8}=\dfrac{19^{20}-8+13}{19^{20}-8}=1+\dfrac{13}{19^{20}-8}\)
\(D=\dfrac{19^{21}+6}{19^{21}-7}=\dfrac{19^{21}-7+13}{19^{21}-7}=1+\dfrac{13}{19^{21}-7}\)
Vì \(19^{20}-8< 19^{21}-7\) và \(13>0\)
\(\Rightarrow\dfrac{13}{19^{20}-8}< \dfrac{13}{19^{21}-7}\)
\(\Rightarrow1+\dfrac{13}{19^{20}-8}< 1+\dfrac{13}{19^{21}-7}\)
\(\Rightarrow\) \(C< D\)
Vậy C < D.
\(M=\dfrac{120-\dfrac{1}{2}.40.5.\dfrac{1}{5}.20.\dfrac{1}{4}-20}{1+5+....+41}\\ =\dfrac{120-20.5-20}{1+5+...+41}\\ =\dfrac{0}{1+5+...+41}\\ =0\)
\(N=10101\left(\dfrac{6}{3.7.11.13.37}+\dfrac{6}{2.3.7.11.13.37}-\dfrac{7}{3.7.11.13.37}\right)\\ =10101\left(-\dfrac{2}{2.3.7.11.13.37}+\dfrac{6}{2.3.7.11.13.37}\right)\\ =3.7.13.37\left(\dfrac{4}{2.3.7.11.13.37}\right)=\dfrac{4}{2.11}=\dfrac{2}{11}\)
bài 2:để Z là số nguyên thì 3n-5 \(⋮\)n+4
\(\Rightarrow[(3n-5)-3(n+4)]⋮(n+4)\)
\(\Rightarrow(3n-5-3n-12)⋮(n+4)\)
\(\Rightarrow-17⋮n+4\)
\(\Rightarrow n+4\inƯ(17)\)={1;-1;17;-17}
\(\Rightarrow\)n\(\in\){-3;-5;13;-21}
a, Ta có
\(\dfrac{111}{37}< x< \dfrac{91}{37}\Leftrightarrow3< x< 7\Leftrightarrow x\in\left\{4,5,6\right\}\)
b,Ta co
\(\dfrac{-84}{14}< 3x< \dfrac{108}{9}\Leftrightarrow3.\dfrac{-28}{14}< 3x< 3.\dfrac{36}{9}\Leftrightarrow\dfrac{-28}{14}< x< \dfrac{36}{9}\Leftrightarrow-2< x< 4\Leftrightarrow x\in\left\{-1,0,1,2,3\right\}\)
Mấy bài dễ u tự giải quyết nha
3) \(\dfrac{2013}{2014}+\dfrac{2014}{2015}+\dfrac{2015}{2013}\)
\(=\left(1-\dfrac{1}{2014}\right)+\left(1-\dfrac{1}{2015}\right)+\left(1+\dfrac{2}{2013}\right)\)
\(=3+\dfrac{2}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\)
\(=3+\left(\dfrac{1}{2013}-\dfrac{1}{2014}\right)+\left(\dfrac{1}{2013}-\dfrac{1}{2015}\right)>3\)
ta thấy:
\(A=\dfrac{37^{20}}{37^{20}-6}< \dfrac{37^{20}+4}{37^{20}-6}< \dfrac{37^{20}+4}{37^{20}-2}\)
\(\Rightarrow A< \dfrac{37^{20}+4}{37^{20}-2}\)
vậy...