LL
1
K
Khách
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LL
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
31 tháng 5
Ta có: \(3A=\frac{3^{19}+6}{3^{19}+10}=\frac{3^{19}+10-4}{3^{19}+10}=1-\frac{4}{3^{19}+10}\)
\(3B=\frac{3^{18}+6}{3^{18}+10}=\frac{3^{18}+10-4}{3^{18}+10}=1-\frac{4}{3^{18}+10}\)
Ta có: \(3^{19}+10>3^{18}+10\)
=>\(\frac{4}{3^{19}+10}<\frac{4}{3^{18}+10}\)
=>\(\frac{-4}{3^{19}+10}>\frac{-4}{3^{18}+10}\)
=>\(\frac{-4}{3^{19}+10}+1>\frac{-4}{3^{18}+10}+1\)
=>3A>3B
=>A>B
LL
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
31 tháng 5
Ta có: \(3A=\frac{3^{19}+6}{3^{19}+10}=\frac{3^{19}+10-4}{3^{19}+10}=1-\frac{4}{3^{19}+10}\)
\(3B=\frac{3^{18}+6}{3^{18}+10}=\frac{3^{18}+10-4}{3^{18}+10}=1-\frac{4}{3^{18}+10}\)
Ta có: \(3^{19}+10>3^{18}+10\)
=>\(\frac{4}{3^{19}+10}<\frac{4}{3^{18}+10}\)
=>\(\frac{-4}{3^{19}+10}>\frac{-4}{3^{18}+10}\)
=>\(\frac{-4}{3^{19}+10}+1>\frac{-4}{3^{18}+10}+1\)
=>3A>3B
=>A>B
LL
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
31 tháng 5
Ta có: \(3A=\frac{3^{19}+6}{3^{19}+10}=\frac{3^{19}+10-4}{3^{19}+10}=1-\frac{4}{3^{19}+10}\)
\(3B=\frac{3^{18}+6}{3^{18}+10}=\frac{3^{18}+10-4}{3^{18}+10}=1-\frac{4}{3^{18}+10}\)
Ta có: \(3^{19}+10>3^{18}+10\)
=>\(\frac{4}{3^{19}+10}<\frac{4}{3^{18}+10}\)
=>\(\frac{-4}{3^{19}+10}>\frac{-4}{3^{18}+10}\)
=>\(\frac{-4}{3^{19}+10}+1>\frac{-4}{3^{18}+10}+1\)
=>3A>3B
=>A>B
LL
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
31 tháng 5
Ta có: \(3A=\frac{3^{19}+6}{3^{19}+10}=\frac{3^{19}+10-4}{3^{19}+10}=1-\frac{4}{3^{19}+10}\)
\(3B=\frac{3^{18}+6}{3^{18}+10}=\frac{3^{18}+10-4}{3^{18}+10}=1-\frac{4}{3^{18}+10}\)
Ta có: \(3^{19}+10>3^{18}+10\)
=>\(\frac{4}{3^{19}+10}<\frac{4}{3^{18}+10}\)
=>\(\frac{-4}{3^{19}+10}>\frac{-4}{3^{18}+10}\)
=>\(\frac{-4}{3^{19}+10}+1>\frac{-4}{3^{18}+10}+1\)
=>3A>3B
=>A>B
LL
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
30 tháng 5
Ta có; \(3A=\frac{3^{19}+6}{3^{19}+10}=\frac{3^{19}+10-4}{3^{19}+10}=1-\frac{4}{3^{19}+10}\)
\(3B=\frac{3^{18}+6}{3^{18}+10}=\frac{3^{18}+10-4}{3^{18}+10}=1-\frac{4}{3^{18}+10}\)
Ta có: \(3^{19}+10>3^{18}+10\)
=>\(\frac{4}{3^{19}+10}<\frac{4}{3^{18}+10}\)
=>\(-\frac{4}{3^{19}+10}>-\frac{4}{3^{18}+10}\)
=>\(-\frac{4}{3^{19}+10}+1>-\frac{4}{3^{18}+10}+1\)
=>3A>3B
=>A>B
LL
1
MH
19 tháng 9 2021
\(A\)\(=\dfrac{3^{18}+2}{3^{19}+10}\)
\(17A=\dfrac{3^{19}+6}{3^{19}+10}=1-\dfrac{4}{3^{19}+10}\)
\(B=\dfrac{3^{17}+2}{3^{18}+10}\)
\(17B=\dfrac{3^{18}+6}{3^{18}+10}=1-\dfrac{4}{3^{18}+10}\)
Vì \(\dfrac{4}{3^{19}+10}< \dfrac{4}{3^{18}+10}\)
⇒\(1-\dfrac{4}{3^{19}+10}\) \(>\) \(1-\dfrac{4}{3^{18}+10}\)
⇒\(A>B\)
Ta có: \(3A=\frac{3^{19}+6}{3^{19}+10}=\frac{3^{19}+10-4}{3^{19}+10}=1-\frac{4}{3^{19}+10}\)
\(3B=\frac{3^{18}+6}{3^{18}+10}=\frac{3^{18}+10-4}{3^{18}+10}=1-\frac{4}{3^{18}+10}\)
Ta có: \(3^{19}+10>3^{18}+10\)
=>\(\frac{4}{3^{19}+10}<\frac{4}{3^{18}+10}\)
=>\(\frac{-4}{3^{19}+10}>\frac{-4}{3^{18}+10}\)
=>\(\frac{-4}{3^{19}+10}+1>\frac{-4}{3^{18}+10}+1\)
=>3A>3B
=>A>B