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H9
HT.Phong (9A5)
CTVHS
13 tháng 1
Tính chất nếu:
\(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\)
Ta có:
\(A=\dfrac{10^{99}+1}{10^{89}+1}>\dfrac{10^{99}+1+9}{10^{89}+1+9}\)
\(A>\dfrac{10^{99}+10}{10^{89}+10}\)
\(A>\dfrac{10\cdot\left(10^{98}+1\right)}{10\cdot\left(10^{88}+1\right)}\)
\(A>\dfrac{10^{98}+1}{10^{88}+1}\)
\(A>B\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1
\(A=\dfrac{10^{99}+1}{10^{89}+1}< \dfrac{10^{99}+1+9}{10^{89}+1+9}=\dfrac{10^{99}+10}{10^{89}+10}=\dfrac{10\left(10^{98}+1\right)}{10\left(10^{88}+1\right)}=\dfrac{10^{98}+1}{10^{88}+1}\)
Vậy \(A< B\)
S
7 tháng 5 2017
a, Ta có: \(\frac{2001}{2002}=\frac{2002-1}{2002}=\frac{2002}{2002}-\frac{1}{2002}=1-\frac{1}{2002}\)
\(\frac{2000}{2001}=\frac{2001-1}{2001}=\frac{2001}{2001}-\frac{1}{2001}=1-\frac{1}{2001}\)
Vì \(\frac{1}{2002}< \frac{1}{2001}\Rightarrow1-\frac{1}{2002}>1-\frac{1}{2001}\Rightarrow\frac{2001}{2002}>\frac{2000}{2001}\)
b, Ta có: \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\Rightarrow\left(\frac{1}{81}\right)^7>\left(\frac{1}{243}\right)^6\Rightarrow\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
c, Ta có: \(\left(\frac{3}{8}\right)^5=\frac{3^5}{\left(2^3\right)^5}=\frac{243}{2^{15}}>\frac{243}{3^{15}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\frac{5^3}{243^3}=\left(\frac{5}{243}\right)^3\)
Vậy \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
d, Ta có: \(\frac{2011}{2012}>\frac{2011}{2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2012+2013}\)
\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
e, \(C=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(D=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)
Vì \(\frac{2}{10^{10}-1}< \frac{2}{10^{10}-3}\Rightarrow1+\frac{2}{10^{10}-1}< 1+\frac{2}{10^{10}-3}\Rightarrow C< D\)
g, \(G=\frac{10^{100}+2}{10^{100}-1}=\frac{10^{100}-1+3}{10^{100}-1}=\frac{10^{100}-1}{10^{100}-1}+\frac{3}{10^{100}-1}=1+\frac{3}{10^{100}-1}\)
\(H=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì \(\frac{3}{10^{100}-1}< \frac{3}{10^8-3}\Rightarrow1+\frac{3}{10^{100}-1}< 1+\frac{3}{10^8-3}\Rightarrow G< H\)
h, Vì E < 1 nên:
\(E=\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=F\)
Vậy E = F
KS
1
30 tháng 3 2020
Ta có A = \(\frac{10^{100}-1}{10^{98}-1}=\frac{10^{98}.10^2-10^2+99}{10^{98}-1}\)
\(=\frac{10^2\left(10^{98}-1\right)+99}{10^{98-1}}\)
\(=10^2+\frac{99}{10^{98}-1}\)
B= \(\frac{10^{101}-1}{10^{99}-1}=\frac{10^{99}.10^2-10^2+99}{10^{99}-1}\)
\(=\frac{10^2\left(10^{99}-1\right)+99}{10^{99}-1}\)
\(=10^2+\frac{99}{10^{99}-1}\)
Vì \(\frac{99}{10^{98}-1}>\frac{99}{10^{99}-1}\)nên \(10^2+\frac{99}{10^{98}-1}>10^2+\frac{99}{10^{99}-1}\)=> A > B
Vậy A > B
30 tháng 4 2019
Bài làm
a ) \(A=\frac{9^{99}+1}{9^{100}+1}=\frac{9^{100}+1}{9^{100}+1}-\frac{9}{9^{100}+1}\)
= \(1-\frac{9}{9^{100}+1}\)
\(B=\frac{10^{98}-1}{10^{99}-1}=\frac{10^{99}-1}{10^{99}-1}-\frac{10}{10^{99}-1}\)
= \(1-\frac{10}{10^{99}-1}\)
Vì \(\frac{9}{9^{100}+1}>\frac{10}{10^{99}-1}\)
nên \(1-\frac{9}{9^{100}+1}< 1-\frac{10}{10^{99}-1}\)
\(\Rightarrow A< B\)
30 tháng 4 2019
Bài làm
b ) \(A=\frac{5^{10}}{1+5+5^2+.....+5^9}=\frac{1+5+5^2+.....+5^9}{1+5+5^2+.....+5^9}+\frac{1+5+5^2+.....+5^8-5^9.4}{1+5+5^2+.....+5^9}\)
= \(1+\frac{1+5+5^2+.....+5^8+5^9.4}{1+5+5^2+.....+5^9}=1+5^9.3\)
\(B=\frac{6^{10}}{1+6+6^2+.....+6^9}=\frac{1+6+6^2+.....+6^9}{1+6+6^2+.....+6^9}+\frac{1+6+6^2+.....+6^8+6^9.5}{1+6+6^2+.....+6^9}\)
= \(1+\frac{1+6+6^2+.....+6^8+6^9.5}{1+6+6^2+.....+6^9}=1+6^9.4\)
Vì \(1+5^9.3< 1+6^9.4\)
nên A < B
LN
0
AH
Akai Haruma
Giáo viên
4 tháng 10
1/
\(10A=\frac{10^{12}-10}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 1< 10B$
$\Rightarrow A< B$
AH
Akai Haruma
Giáo viên
4 tháng 10
2/
\(C=\frac{10^{99}+5}{10^{99}-8}=1+\frac{13}{10^{99}-8}\)
\(D=\frac{10^{100}+6}{10^{100}-4}=1+\frac{10}{10^{100}-4}\)
So sánh \(\frac{13}{10^{99}-8}=\frac{130}{10^{100}-80}> \frac{130}{10^{100}-4}> \frac{10}{100^{100}-4}\)
$\Rightarrow 1+\frac{13}{10^{99}-8}> 1+\frac{10}{100^{10}-4}$
$\Rightarrow C> D$
PN
10 tháng 5 2021
a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)
b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)