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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


a) Ta có:
\(0,3\left(2\right)=0,3222...=0,32+0,00222...\)
Mà: \(0,32+0,00222...>0,32\)
\(\Rightarrow0,3\left(2\right)>0,32\)
b) Ta có:
\(\dfrac{5}{6}=0,8\left(3\right)=0,8333...=0,8+0,0333...\)
\(0,834=0,8+0,034\)
Mà: \(0,0333...< 0,34\)
Nên: \(\dfrac{5}{6}< 0,834\)
a: 0,3(2)=0,3222...>0,32
b: 5/6=0,8(3)=0,83333...<0,834
a) \(0,3\left(2\right)=0,3+\dfrac{2}{90}=\dfrac{29}{90}=\dfrac{145}{450}\)
\(0,32=\dfrac{8}{25}=\dfrac{144}{450}\)
Do \(145>144\Rightarrow\dfrac{145}{450}>\dfrac{144}{450}\)
Vậy \(0,3\left(2\right)>0,32\)
b) \(0,834=\dfrac{417}{500}=\dfrac{1251}{1500}\)
\(\dfrac{5}{6}=\dfrac{1250}{1500}\)
Do \(1251>1250\Rightarrow\dfrac{1251}{1500}>\dfrac{1250}{1500}\)
\(\Rightarrow0,834>\dfrac{5}{6}\)