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1) Phân tích A ra :
A= 1717.17+\(\frac{1}{17^{18}.17}\)+1 So sánh với B ta có: A có 1718>1717 của B nhưng B lại có 1/1718>1/1719.
Mà 1718>1/1718 nên suy ra A>B
2) Bài nay tương tự bài trên.
2/(2012+2013) < 2/(2012 + 2012) = 2/ (2.2012) = 1/2012
2009/(2012+2013) < 2009/2012
=> 2011/(2012+2013) = 2/(2012+2013) + 2009/(2012+2013) < 1/2012 + 2009/2012
=> 2011/(2012+2013) < 2010/2012 (a)
2012/(2012+2013) < 2012/2013 (b)
lấy (a) + (b) => (2011+2012)/(2012+2013) < 2010/2012 + 2012/2013
vậy B < A
Ta có: \(A=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19A=\frac{19.\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
\(B=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19B=\frac{19.\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
Nên \(19A< 19B\Rightarrow A< B\)
Nhầm: Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow A>B\)
C = 1930+5/1931+5
=>19C = 1931+95/1931+5 = 1+ [90/1931+5]
D = 1931+5/1932+5
=>19D = 1932+95/1932+5 = 1 + [90/1932+5]
ma 90/1931+5 > 90/1932+5
=>19C > 19D
=>C > D
Ta có:+) \(A=\frac{2^{19}-3}{2^{20}-3}\)
\(2A=\frac{2^{20}-6}{2^{20}-3}=\frac{\left(2^{20}-3\right)-3}{2^{20}-3}\)
\(2A=1-\frac{3}{2^{20}-3}\)
+)\(B=\frac{2^{20}-3}{2^{21}-3}\)
\(2B=\frac{2^{21}-6}{2^{21}-3}=\frac{\left(2^{21}-3\right)-3}{2^{21}-3}\)
\(2B=1-\frac{3}{2^{21}-3}\)
Vì \(2^{20}-3< 2^{21}-3\)nên \(1-\frac{3}{2^{20}-3}< 1-\frac{3}{2^{21}-3}\)
Hok tốt nha^^
19A= \(\frac{19^{31}+95}{19^{31}+5}\) = \(\frac{19^{31}+95}{19^{31}}\)+\(\frac{19^{31}+95}{5}\)=95+ \(\frac{19^{31}+95}{5}\)
19B=\(\frac{19^{32}+95}{19^{32}+5}\)=\(\frac{19^{32}+95}{19^{32}}\)+\(\frac{19^{32}+95}{5}\)=95+\(\frac{19^{32}+95}{5}\)
vì \(\frac{19^{31}+95}{5}\)<\(\frac{19^{32}+95}{5}\)=> 19A<19B => A<B
Ta có:
\(A=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19A=\frac{19\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
\(B=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19B=\frac{19\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow19A>19B\Rightarrow A>B\)
Ta có: \(A=\frac{17^{18}+1}{17^{19}+1}<1\)
\(A=\frac{17^{18}+1}{17^{19}+1}<\frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=B\)
=> A<B
Để so sánh A =1718+1/1719+1 và B=1717+1/1718+1
=>Ta xét bài toán phụ sau
a/b<1 thì a/b<a+/b+m
vì a/b<1=>a<b mà m thuộc N*
=>a.m<b.m=>ab+am<ab+bm
a/b=a.(b+m0/b.(b+m)/b(b+m=ab+am/b(b+m)<ab+bm/b(b+m)
Vì b(b+m)>0=>a/b<ab+bm/b(b+m)=b(a+m)/b(b+m)=a+m/b+m
=>.a/b<a+m/b+m(1)
vì 1718+ 1 < 1719+1
=>A<1
(1)=>1718+1/1719+1<1718+1+16/1719+1+16
<=>A<1717+17/1719+17=17(1717+1)/1791718+1)
<=>A<1717+1/1718+1=B
<=>A<B
Vậy...
\(Q=\frac{2010+2011+2012}{2011+2012+2013}\)
\(Q=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
Ta có :
\(\hept{\begin{cases}\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\\\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\\\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\end{cases}}\)
\(\Rightarrow P>Q\)