A=1999mũ 1999 +1 / 1999mũ1998 +1
và B=1999mũ2000 +1 / 1999mũ1999
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a) Do A = 98 99 + 1 98 89 + 1 > 1 nên
A = 98 99 + 1 98 89 + 1 > 98 99 + 1 + 97 98 89 + 1 + 97 = 98 ( 98 98 + 1 ) 98 ( 98 88 + 1 ) = 98 98 + 1 98 88 + 1 = B
Vậy A > B
b) Do C = 100 2008 + 1 100 2018 + 1 < 1 nên
C= 100 2008 + 1 100 2018 + 1 > 100 2008 + 1 + 99 100 2018 + 1 + 99 = 100 ( 100 2007 + 1 ) 100 ( 100 2017 + 1 ) = 100 2007 + 1 100 2017 + 1 = D
Vậy C > D.
A) x/y-3/8=1 x/y-3/8=1/2 B)4/9:x/y=1 4/9:x/y=2/3
x/y=1+3/8 x/y=1/2+3/8 x/y=4/9:1 x/y=4/9:2/3
x/y=8/8+3/8 x/y=4/8+3/8 x/y=4/9 x/y=2/3
x/y=11/8 x/y=7/8
\(1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}\times1\frac{1}{5}\times...1\frac{1}{9}+2005.\)
\(=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\frac{6}{5}\times...\times\frac{10}{9}+2005\)
\(=\frac{10}{2}+2005=5+2005=2010\)
.
Áp dụng BDT svacxơ ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=\frac{9}{1}=9\)
Vì \(a+b+c=1\)
Dấu ''='' khi a=b=c
Học tốt.
\(\frac{A}{2}=\frac{2^{2015}+1}{2\left(2^{2014}+1\right)}=\frac{2^{2015}+1}{2^{2015}+2}=\frac{2^{2015}+2-1}{2^{2015}+2}=1-\frac{1}{2^{2015}+2}\)
\(\frac{A}{2}=\frac{2^{2016}+1}{2\left(2^{2015}+1\right)}=\frac{2^{2016}+1}{2^{2016}+2}=\frac{2^{2016}+2-1}{2^{2016}+2}=1-\frac{1}{2^{2016}+}\)
Vì \(1-\frac{1}{2^{2015}+2}< 1-\frac{1}{2^{2016}+2}\Rightarrow\frac{A}{2}< \frac{B}{2}\)
\(\Rightarrow A< B\)
\(1\frac{1}{15}\cdot1\frac{1}{16}\cdot1\frac{1}{17}\cdot...\cdot1\frac{1}{2006}\)
\(=\frac{16}{15}\cdot\frac{17}{16}\cdot\frac{18}{17}\cdot...\cdot\frac{2007}{2006}\)
\(=\frac{16\cdot17\cdot18\cdot...\cdot2007}{15\cdot16\cdot17\cdot...\cdot2006}\)
\(=\frac{2007}{15}\)
1\(\frac{1}{15}\) x 1\(\frac{1}{16}\) x 1\(\frac{1}{17}\) x ............ x 1\(\frac{1}{2016}\)
= \(\frac{16}{15}\)x \(\frac{17}{16}\)x \(\frac{18}{17}\)x ................. x \(\frac{2017}{2016}\)
= \(\frac{1}{15}\)x 2017
= \(\frac{2017}{15}\)
SO SÁNH Avà B á hay tính z bn
#)Giải :
Ta có :
\(A=\frac{1999^{1999}+1}{1999^{1998}+1}=\frac{1999^{1999}+1999-1998}{1999^{1998}+1}=1999-\frac{1998}{1999^{1998}+1}\)
\(B=\frac{1999^{2000}+1}{1999^{1999}+1}=\frac{1999^{2000}+1999-1998}{1999^{1999}+1}=1999-\frac{1998}{1999^{1999}+1}\)
Vì \(1999^{1998}+1< 1999^{1999}+1\)
\(\Rightarrow\frac{1}{1999^{1998}+1}>\frac{1}{1999^{1999}+1}\Rightarrow1999+\frac{-1}{1999^{1998}+1}< 1999+\frac{-1}{1999^{1999}+1}\Rightarrow A< B\)