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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
\(5^{200}=\left(5^2\right)^{100}=25^{100}\)
\(3< 25=>3^{100}< 25^{100}=>3^{100}< 5^{200}\)
\(\frac{75^{20}}{45^{10}.25^{15}}=\frac{25^{20}.3^{20}}{3^{10}.3^{10}.5^{10}.25^{15}}=\frac{25^{20}}{25^5.25^{15}}=1\)
\(=>75^{20}=45^{10}.25^{15}\left(dpcm\right)\)
P/S:nếu a=b=>a:b=1 mk làm theo cách đó cho nhanh mà bn ghi sai đề r
1.
a)=1/3-[(-5/4)-5/8]
=1/3-(-15/8)=53/24
b)=5/9:(-3/22)+5/9:(-3/5)
=5/9*22/-3+5/9*5/-3=-110/27+-25/27=5
2
a)Ta có 339<340=920<1120<1121
nên 339<1121
b)Ta có /3,4-x/ lớn hơn hoặc bằng 0 Với mọi x thuộc R
=> -/3,4-x/ bé hơn hoặc bằng 0 Với mọi x thuộc R
=> 0,5-/3,4-x/ bé hơn hoặc bằng 0,5 Với mọi x thuộc R
Dấu = xảy ra khi 3,4-x=0
=>x=3,4
Vậy GTLN của A = 0,5 khi x=3,4
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