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Ta co
+)A=2/60*63+2/63*66+...+2/117*120+2/2003
A*3/2=3/60*63+3/63*66+...+3/117*120+3/2003
A*3/2=1/60-1/63+1/63-1/66+...+1/117-1/120+3/2003
A*3/2=1/60-1/120+3/2003
A=(1/120+3/2003)*2/3
+)B=5/40*44+5/44*48+...+5/76*80+5/2003
B*4/5=4/40*44+4/44*48+...+4/76*80+4/2003
B*4/5=1/40-1/44+1/44-1/48+...+1/76-1/80+4/2003
B*4/5=1/40-1/80+4/2003
B=(1/80+4/2003)*5/4
Tu tren ta co A=(1/120+3/2003)*2/3
B=(1/80+4/2003)*5/4
Vay A<B(Vi 1/120<1/80;3/2003<4/2003;2/3<5/4)
Ta có: \(A=\frac{2}{60.63}+\frac{2}{63.66}+...+\frac{2}{117.120}+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{3}{60.63}+\frac{3}{63.66}+...+\frac{3}{117.120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{63}+\frac{1}{63}-\frac{1}{66}+...+\frac{1}{117}-\frac{1}{120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}\left(\frac{1}{60}-\frac{1}{120}\right)+\frac{2}{2003}\)
\(\Rightarrow A=\frac{2}{3}.\frac{1}{120}+\frac{2}{2003}\)
\(\Rightarrow A=\frac{1}{180}+\frac{2}{2003}\)
\(B=\frac{5}{40.44}+\frac{5}{44.48}+...+\frac{5}{76.80}+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{4}{40.44}+\frac{4}{44.48}+...+\frac{4}{76.80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{44}+\frac{1}{44}-\frac{1}{48}+...+\frac{1}{76}-\frac{1}{80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}\left(\frac{1}{40}-\frac{1}{80}\right)+\frac{5}{2003}\)
\(\Rightarrow B=\frac{5}{4}.\frac{1}{80}+\frac{5}{2003}\)
\(\Rightarrow B=\frac{1}{64}+\frac{5}{2003}\)
Vì \(\left\{\begin{matrix}\frac{1}{64}>\frac{1}{180}\\\frac{5}{2003}>\frac{2}{2003}\end{matrix}\right.\Rightarrow\frac{1}{64}+\frac{5}{2003}>\frac{1}{180}+\frac{2}{2003}\Rightarrow B>A\)
Vậy A < B
a) \(\sqrt{3}+5=\sqrt{3}+\sqrt{25}>\sqrt{2}+\sqrt{11}\)
b) \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
c) \(4+\sqrt{33}=\sqrt{16}+\sqrt{33}>\sqrt{29}+\sqrt{14}\)
d) \(\sqrt{48}+\sqrt{120}< \sqrt{49}+\sqrt{121}=7+11=18\)
a )
Ta có :
\(3^{40}=\left(3^4\right)^{10}=81^{10}\)
\(5^{30}=\left(5^3\right)^{10}=125^{10}\)
Do \(125^{10}>81^{10}\)
\(\Rightarrow5^{30}>3^{40}\)
b )
Ta có : \(5^{303}>2^{44}\left(5>2;303>44\right)\)
c )
Ta có : \(5^{303}>2^4\left(5>2;303>4\right)\)
a) 340 = (33)10 = 910
530 = ( 53 )10 = 12510
Mà \(9^{10}< 125^{10}\Rightarrow3^{40}< 5^{30}\)
Vậy ....
b) 5303 > 244 ( vì 5 > 2 ; 303 > 44 )
c) 5303 > 24 ( vì 5 > 2 ; 303 > 4 )
a -35/50 = -7/10
b 510/2805 = 2/11
c 119/126
B2
-2/3= -8/12 , -1/4= -3/12
-8/12<-3/12 nên -2/3<-1/4
b 2/3 5/6
12/18 và 15/18
12/18<15/18
nên 14/21<60/72
b: \(A=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2003}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{2}{2003}\)
\(=2\left(\dfrac{1}{360}+\dfrac{1}{2003}\right)\)
\(B=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}\right)+\dfrac{5}{2003}\)
\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2003}\)
\(=5\left(\dfrac{1}{320}+\dfrac{1}{2003}\right)\)
Vì 1/360+1/2003<1/320+1/2003
nên A<B