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\(1-\frac{1003}{1005}=\frac{2}{1005}>\frac{2}{1007}=1-\frac{1005}{1007}\Rightarrow\frac{1003}{1005}<\frac{1005}{1007}\)
ta có : 1-1003/1005=2/1005
1-1005/1007=2/1007
vì 2/1005>2/1007 nên 1003/1005<1005/1007
Ta có: B=\(\frac{17^{2009}+1}{17^{2010}+1}\)<1 ( Vì 172009+1< 172010+1 )
Nên B=\(\frac{17^{2009}+1}{17^{2010}+1}\)<\(\frac{17^{2009}+1+16}{17^{2010}+1+16}\)
=\(\frac{17^{2009}+17}{17^{2010}+17}\)
=\(\frac{17\left(17^{2008}+1\right)}{17\left(17^{2009}+1\right)}\)
=\(\frac{17^{2008+1}}{17^{2009}+1}\)=A
Vậy A>B
a)\(x.3^{15}=3^{17}\)
\(x=3^{17}:3^{15}\)
\(x=3^2=9\)
b) \(5^x=6^x\Leftrightarrow x=1;x=0\)
c) \(x^3=x^6\)
\(x^3=x^3.x^3\) \(x^3=1\) \(x=1\) | \(x^3=\left(x^3\right)^2\) \(x=0\) |
B2 ss
a)\(3^{45}=\left(3^3\right)^{15}=27^{15}\)
\(4^{30}=\left(4^2\right)^{15}=16^{15}\)
vì 1615 < 2715 nên 430 < 345
b)
\(818.820=\left(819-1\right)\left(819+1\right)=819^2-1\)
vì 8192 > 8192 - 1 nên 8192 > 818.820
\(B=3^2+3^3+...+3^{99}\)
\(3B=3^3+3^4+...+3^{100}\)
\(3B-B=\left(3^3+3^4+...+3^{100}\right)-\left(3^2+3^3+...+3^{99}\right)\)
\(2B=3^{100}-3^2\)
\(B=\frac{3^{100}-9}{2}\)
\(2B+9=3^{2n+4}\)
\(\Leftrightarrow3^{2n+4}=3^{100}\)
\(\Leftrightarrow2n+4=100\)
\(\Leftrightarrow n=48\).
\(\dfrac{4}{17}=\dfrac{16}{68}\\ Vì:\dfrac{16}{68}< \dfrac{16}{63}\Rightarrow\dfrac{4}{17}< \dfrac{16}{63}\\ ---\\ \dfrac{1007}{1009}=1-\dfrac{2}{1009};\dfrac{1005}{1007}=1-\dfrac{2}{1007}\\ Vì:\dfrac{2}{1009}< \dfrac{2}{1007}\Rightarrow1-\dfrac{2}{1009}>1-\dfrac{2}{1007}\\ \Rightarrow\dfrac{1007}{1009}>\dfrac{1005}{1007}\)
a: 4/17=16/68
16/68<16/63
=>4/17<16/63
b: 19/53<20/53
20/53<20/50(Vì 53>50)
=>19/53<20/50=2/5
mà 2/5=30/75<30/73
nên 19/53<30/73
c: 1007/1009=1-2/1009
1005/1007=1-2/1007
1009>1007
=>2/1009<2/1007
=>-2/1009>-2/1007
=>1007/1009>1005/1007